ALBERT R. MANN LIBRARY New York State Colleges OF " Agriculture and Home Economics AT Cornell University Date Due Uhi 12 AUG 10 -■»%^^ ^ -ilH i ir fe» H8 8 g ' - 13 : 0£C ■Z^p} n -ibrary Bureau Cat. Wfi. 1137 Cornell University Library QA 453.W5 Plane and solid geomrt^.Teach^^^^^^^^^^ b 3 1924 002 961 344 B Cornell University B Library The original of tliis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924002961344 MATHEMATICAL TEXT-BOOKS BY GEORGE A. WENTWORTH Mental Arithmetic Elementaiy Arithmetic Practical Arithmetic Primary Arithmetic (Wentwoith and Eeed) Grammar School Arithmetic Advanced Arithmetic First Steps in Algehra New School Algebra School Algebra Elements of Algebra (Revised Edition) Shorter Course in Algebra Complete Algebra Higher Algebra College Algebra (Revised Edition) First Steps in Geometry (Wentworth and HUl) Plane Geometry (Revised) Solid Geometry (Revised) Plane and SoUd Geometry (Revised) Syllabus of Geometry Geometrical Exercises Analytic Geometry Plane and Solid Geometry and Plane Trigonometry (Second Revised Edition) Plane Trigonometry (Second Revised Edition) Plane Trigonometry and Tables (Second Revised Edition) Plane and Spherical Trigonometry (Second Revised Edition) Plane and Spherical Trigonometry and Tables (Second Revised Edition) Plane and Spherical Trigonometry, Surveying, and Navigation (Second Revised Edition) Surveying and Tables (Second Revised Edition) Plane Trigonometry, Surveying, and Tables (Second Revised Edition) Plane and Spherical Trigonometry, Surveying, and Tables (Second Revised Edition) Logarithms, Metric Measures, etc. PLANK AND SOLID GEOMBTET G. A. V^ENTWOETH AtTTHOE OF A SEBIES OF TEXT-BOOKS Df MATJiEMLiTICS REVISED EDITION GINN & COMPANY BOSTON • NEW TOEK ■ CHICAGO • LONDON £,V, H Entered, according to Act of Congress, in the year 1888, by G. A. WENXWOETH in tlie Office of tlie Librarian of Congress, at Washington oopykight, 1899 By G. a. WENTWOETH JiTT. RIGHTS KESERVED 814.1 Vf)t gtjienaum jBtesa GINN & COMPANY ■ PRO. PRIETOKS ■ BOSTON ' U.S.A. PREFACE. Most persons do not possess, and do not easily acquire, the power of atstraction requisite for apprehending geometrical conceptions, and for keeping in mind the successive steps of a continuous argument. Hence, with a very large proportion of beginners in Geometry, it depends mainly upon the form in which the subject is presented whether they pursue the study with indifference, not to say aversion, or with increasing interest and pleasure. Great care, therefore, has been taken to make the pages attractive. The iigm'es have been carefully drawn and placed in the middle of the page, so that they fall directly under the eye in immediate connec- tion with the text; and in no case is it necessary to turn the page in reading a demonstration. Full, long-dashed, and short^dashed lines of the figures indicate given, resulting, and auxiliary lines, respectively. Bold-faced, italic, and roman type has been skilfully used to distinguish the hypothesis, the conclusion to be proved, and the proof. As a further concession to the beginner, the reason for each statement in the early proofs is printed ui small italics, immediately following the statement. This prevents the necessity of interrupting the logical train of thought by turning to a previous section, and compels the learner to become familiar with a large number of geometrical truths by constantly seeing and repeating them. This help is gradually discarded, and the pupil is left to depend upon the knowledge already acquired, or to find the reason for a step by tiirning to the given reference. It must not be inferred, because this is not a geometry of interrogation points, that the author has lost sight of the real object of the study. The training to be obtained from carefully following the logical steps of a complete proof has been provided for by the Propositions of the IV PREFACE. Geometry, and the development of the power to grasp and prove new truths has been provided for by original exercises. The chief value of any Geometry consists in the happy combination of these two kinds of training. The exercises have been arranged according to the test of experience, and are so abundant that it is not expected that any one class will work them all out. The methods of attacking and prov- ing original theorems are fully explained in the first Book, and illus- trated by sufficient examples ; and the methods of attackiug and solving origmal problems are explained in the second Book, and illustrated by examples worked out in full. None but the very simplest exercises are inserted until the student has become familiar with geometrical methods, and is furnished with elementary but much needed instruction in the art of handling original propositions ; and he is assisted by diagrams and hints as long as these helps are necessary to develop his mental powers sufficiently to enable him to carry on the work by himself. The law of converse theorems, the distinction between positive and negative quantities, and the principles of reciprocity and continuity have been briefly explained; but the application of these principles is left mainly to the discretion of teachers. The author desires to express his appreciation of the valuable sugges- tions and assistance which he has received from distinguished educators in all parts of the country. He also desires to acknowledge his obliga/- tion to Mr. Charles Hamilton, the Superintendent of the composition room of the Athenaeum Press, and to Mr. I. P. White, the compositor, for the excellent typography of the book. Criticisms and corrections will be thankfully received. G. A. WENTWORTH. Exeter, N. H., June, 1899. NOTE TO TEACHERS. It is intended to have the first fourteen pages of this book simply read in the class, with such running comment and discussion as may be useful to help the beginner catch the spirit of the subject-matter, "and not leave him to the mere letter of dry definitions. In like manner, the definitions at the beginning of each Book should be read and discussed in the recitation room. There is a decided advantage in having the definitions for each Book in a single group so that they can be included in one survey and discussion. For a similar reason the theorems of limits are considered together. The subject of limits is exceedingly interesting in itself, and it was thought best to include in the theory of limits in the second Book every principle required for Plane and Solid Geometry. "When the pupil is reading each Book for the first time, it will be well to let him write his proofs on the blackboard in his own lan- guage, care being taken that his language be the simplest possible, that the arrangement of work be vertical, and that the figures be accurately constructed. This method will furnish a, valuable exerciise as a language lesson, will cultivate the habit of neat and orderly arrangement of work, and will allow a brief interval for deliberating on each step. After a Book has been read in this way, the pupil should review the Book, and should be required to draw the figures free-hand. He should state and prove the propositions orally, using a pointer to indi- cate on the figure every line and angle named. He should be encour- aged, in reviewing each Book, to do the original exercises ; to state the converse propositions, and determine whether they are true or false ; and also to give well-considered answers to questions which may be asked him on many propositions. V VI NOTE TO TEACHERS. The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus, a rectangle with a con- stant base 6, and a variable altitude x, will afford an obvious illus- tration of the truth that the product of a constant and a variable is also a variable ; and that the limit of the product of a constant and a variable is the product of the constant by the limit of the variable. If X increases and approaches the altitude a as a limit, the area 6f the rectangle increases and approaches the area of the rectangle a6 as a limit ; if, however, x decreases and approaches zero as a limit, the area of the rectangle decreases and approaches zero as a limit. An arithmetical illustration of this truth may be given by multiplying the approximate values of any repetend by a constant. If, for example, we take the repetend 0.3333 etc., the approximate values of the repe- tend will be j%, jVffi tVo%> i¥!!¥i!' ^"^^ > ^^'^ these values multiplied by 60 give the series 18, 19.8, 19.98, 19.998, etc., which evidently approaches 20 as a limit ; but the product of 60 into | (the limit of the repetend 0.333 etc.) is also 20. Again, if we multiply 60 into the different values of the decreasing series j-'^i jiv jttVt!' jdisiji ^'o-' which approaches zero as a limit, we shall get the decreasing series 2, J, 5V, zh^^ ^^'^•> ^^"^ ^^^ series evi- dently approaches zero as a limit. The Teacher is likewise advised to give frequent written examina- tions. These should not be too difficult, and sufficient time should be allowed for accurately constructing the figures, for choosing the best language, and for determining the best arrangement. The time necessary for the reading of examination books will be diminished by more than one half, if the use of symbols is allowed. Exeter, N.H., 1899. CONTENTS. GEOMETRY. rAOB Introduction . .... ... 1 Genekal Terms . .... .4 General Axioms . 6 Symbols and Abbreviations 6 PLANE GEOMETRY. BOOK I. RECTILINEAR FIGURES. Definitions ..... . .7 The Straight Line . 8 The Plane Angle . .... 9 Extension of the Meaning of Angles 12 Unit of Angles ........ 13 Perpendicdlar and Oblique Lines . , .15 Parallel Lines ...... . . 24 Triangles ....... ... 30 Loci op Points . . 44 Quadrilaterals 47 Polygons in General 56 Symmetry .... 60 Methods op Proving Theorems 64 Exercises 68 vii viu CONTENTS. BOOK II. THE CIRCLE. PAGE Definitions .......... 75 Arcs, Chords, and Tangents 77 Theory of Limits 93 Measure of Angles 100 Exercises . 108 Problems op Construction 112 Solution op Problems ........ 127 Exercises . ..... 129 BOOK III. PROPORTION. SIMILAR POLYGONS. Theory of Proportion 135 Similar Polygons 148 Exercises 159 Numerical Properties of Figures ..... 160 Exercises 170 Problems of Construction ....... 172 Exercises . . . 178 BOOK IV. AREAS OF POLYGONS. Areas of Polygons . . 184 Comparison of Polygons . . ..... 192 Exercises . . 195 Problems of Constrijction , .... 197 Exercises . . . ...... 207 BOOK V. REGULAR POLYGONS AND CIRCLES. Regular Polygons and Circles 211 Problems of Construction 226 Maxima and Minima . 234 Exercises 241 CONTENTS. IX BOOK VI, DEFnriTiONS Lines and Planes Dihedral Angles Polyhedral Angles Exercises . SOLID GEOMETEY. LINES AND PLANES IN SPACE. PAGE 251 253 269 282 2b8 BOOK VIL POLYHEDRONS, CYLINDERS, AND CONES. Polyhedrons 289 Prisms and Parallelopipeds . 290 Exercises 306 Pyramids . 307 General Theorems op Polyhedrons 324 Similar Polyhedrons 326 Regular Polyhedrons . 330 Cylinders 332 Exercises . 341 Cones ... 342 The Prismatoid Formula ...... 354 Exercises 357 BOOK VHL THE SPHERE. Plane Sections and Tangent Planes Figures on the Surface op a Sphere Measurement of Spherical Surfaces Spherical Volumes .... Exercises ..... MiSCliLLANEOUS ExERCISES 360 370 388 397 402 405 X CONTENTS. BOOK IX. CONIC SECTIONS. PAGE The Parabola 409 ExEKOiSES 424 The Ellipse 425 Exercises 443 The Hypekeola 445 TABLE OF FORMULAS 460 INDEX 465 GEOMETRY. 0>«»:;0 INTRODUCTION. 1. If a block of wood or stone is cut in the shape repre- sented in Fig. 1, it will have six flat faces. Each face of the block is called a surface ; and if the faces are made smooth by polishing, so that, when a straight edge is applied to any one of them, the straight edge in every part will touch the surface, the faces are called plane surfaces, or planes. pig. i. 2. The intersection of any two of these surfaces is called a line. 3. The intersection of any three of these lines is called a point. 4. The block extends in three principal directions : From left to right, A to B. From front to back, A to C. From top to bottom, A to _D. These are called the dimensions of the block, and are named in the order given, length, breadth (or width), and thickness (height or depth). 2 GEOMETRY. 5. A solid, in common language, is a limited portion of space JiUed with matter; but in Geometry we have nothing to do with the matter of which a body is composed; we study simply its shape and size; that is, we regard a solid as a limited portion of space which may be occupied by a physical body, or marked out in some other way. Hence, A geometrical solid is a limited portion of space. 6. The surface of a solid is simply the boundary of the solid, that which separates it from surrounding space. The surface is no part of a solid and has no thickness. Hence, A surface has only two dimensions, length and breadth. 7. A line is simply a boundary of a surface, or the inter- section of two surfaces. Since the surfaces have no thickness, a line has no thickness. Moreover, a line i ) no pai't of a surface and has no width. Hence, A line has only one dimension, length. 8. A point is simply the extremity of a line, or the inter- section of two lines. A point, therefore, has no thickness, width, or length; therefore, no magnitude. Hence, A point has no dimension, but denotes position simply. 9. It must be distinctly understood at the outset that the points, lines, surfaces, and solids of Geometry are purely ideal, though they are represented to the eye in a material way. Lines, for example, drawn on paper or on the blackboard, will have some width and some thickness, and will so far fail of being true lines; yet, when they are used to help the mind in reasoning, it is assiimed that they represent true lines, without breadth and without thickness. INTRODUCTION. 3 10. A point is represented to the eye by a fine dot, and named by a letter, as A (Fig. 2). A line is named by two letters, placed one at each end, as BF. A surface is represented and named by / 7 the lines which bound it, as BCDF. A / ^ / solid is represented by the faces which ^ F bound it. pig. 2. 11. A point in space may be considered by itself, without reference to a line. 12. If a point mores in space, its path is a line. This line may be considered apart from the idea of a surface. 13. If a line moves in space, it generates, in general, a surface. A surface can then be considered apart from the idea of a solid. 14. If a surface moves in space, it generates, in general, a solid. D s Thus, let the upright surface ABCD J e/ (Fig. 3) move to the right to the position EFGH, the points A, B, C, and D gener- \Q_ aq ating the lines AE, BF, CG, and DH, ]/_ J/ respectively. The lines AB, BC, CD, and DA will generate the surfaces AF, BG, CB, and DE, respectively. The surface ABCD will generate the solid AG. 15. Geometry is the science which treats of position, form, and maffiiitude. 16. A geometrical figure is a combination of points, lines, surfaces, or solids. 17. Plane Geometry treats of figures all points of which are in the same plane. Solid Geometry treats of figures all points of which are not in the same plane. GEOMETRY. GENERAL TERMS. 18. A proof is a course of reasoning by which the truth or falsity of any statement is logically established. 19. An axiom is a statement admitted to be true without proof. 20. A theorem is a statement to be proved. 21. A construction is the representation of a required figure by means of points and lines. 22. A postulate is a construction admitted to be possible, 23. A problem is a construction to be made so that it shall satisfy certain given conditions. 24. A proposition is an axiom, a theorem, a postulate, or a problem. 25. A corollary is a truth that is easily deduced from known truths. 26. A scholium is a remark upon some particular feature of a proposition. 27. The solution of a problem consists of four parts : 1. The analysis, or course of thought by which the con- struction of the required figure is discovered. 2. The construction of the figure with the aid of ruler and compasses. 3. The proof that the figure satisfies all the conditions. 4. The discussion of the limitations, if any, within which the solution is possible. INTRODUCTION. 5 28. A theorem consists of two pajts : the hypothesis, or that which is assvuned ; and the conclusion, or that which is asserted to follow from the hypothesis. 29. The contradictory of a theorem is a theorem which must be true if the given theorem is false, and must be false if the given theorem is true. Thus, A theorem . If ^ is B, then C is D. Its contradictory : If ^ is B, then C is not D. 30. The opposite of a theorem is obtained by making both the hypothesis and the conclusion negative. Thus, A theorem : If ^ is B, then C is D. Its opposite : If ^ is not B, then C is not Z>. 31. The converse of a theorem is obtained by interchanging the hypothesis and conclusion. Thus, A theorem : If ^ is B, then C is D. Its converse : If C is D, then A is B. 32. The converse of a truth is not necessarily true. Thus, Every horse is a quadruped is true, hut the converse, Every quadruped is a horse, is not true. 33. If a direct proposition and its opposite are true, the converse proposition is true ; and if a direct proposition and its converse are true, the opposite proposition is true. Thus, if it were true that 1. If an animal is a horse, the animal is a quadruped ; ' 2. If an animal is not a horse, the animal is not a quadruped ; '-•' it would follow that 3. If an animal is a quadruped, the animal is a horse. Moreover, if 1 and 3 were true, then 2 would he true. .GEOMETRY. 34. GENERAL AXIOMS. 1. Magnitudes -which are equal to the same magnitude, or equal magnitudes, are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals ai-e taken from equals, the remainders are equal. 4. If equals are added to unequals, the sums are unequal in the same order ; if unequals are added to unequals in the same order, the sums are unequal in that order. 5. If equals are taken from unequals, the remainders are unequal in the same order ; if unequals are taken from equals, the remainders are unequal iu the reverse order. 6. The doubles of the same magnitude, or of equal magni- tudes, are equal ; and the doubles of unequals are unequal, 7. The halves of the same magnitude, or of equal magni tudes, are equal ; and the halves of unequals are unequal. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 35. SYMBOLS AND ABBREVIATIONS. > is (or are) greater than. Def. . . . definition. < is (or are) less tlian. =c= is (or are) equivalent to. .-. therefore. _L perpendicular. Jl perpendiculars. II parallel, lis parallels. Z. angle. A angles. A triangle. A triangles. O parallelogram. m parallelograms. O circle. ® circles, rt. right, st. straight. Q.E.D. stands for quod erat demonstrandum, which was to be proved. Q.E.r. stands for quod erat faciendum, which was to be done. The signs +, — , x , -f-, =, have the same meaning as In Algebra. Ax. . . . axiom. Hyp. . . hypothesis. Cor. . . corollary. Scho. . scholium. Ex. . . exercise. Adj. . . . adjacent; Iden. . identical. Const. . construction. Sup. . supplementary. Ext. . . . exterior. Int. . . . interior. Alt. . . . alternate. PLANE GEOMETRY. Book I. RECTILINEAR FIGURES. DEFINITIONS. 36. A straight line is a line such that any part of it, how- ever placed on any other part, "wUl lie wholly in tjiat part if its extremities lie in that part, as AB. 37. A curved line is a line no part of ^ ^ which is straight, as CD. 38. A broken line is made up of dif- e- Fig. 4. ferent straight lines, as EF. Note. A straight line is often called simply a line. 39. A plane surface, or a plane, is a surface in which, if any two points are taken, the straight line joining these points lies wholly in the surface. 40. A curved surface is a surface no part of which is plane. 41. A plane figure is a figure all points of which are in the same plane. 42. Plane figures which are bounded by straight lines are called rectilinear figures ; by curved lines, curvilinear figures. 43. Figures that have the same shape are called similar. Figures that have the same size but not the same shape are called equivalent. Figures that have the same shape and the same size are called equal or congruent. 7 BOOK I. PLANE GEOMETRY. THE STRAIGHT LIKE. 44. Postulate. A straight line can be drawn from one point to another. 45. Postulate. A straight line can be produced indefinitely. 46. Axiom.* Only one straight line can he drawn from one point to another. Hence, two points determine a straight line. 47. CoR. 1. Tu'o straight lines which have two points in common coincide and form but one line. 48. Cor. 2. Two straight lines can intersect in only one point. For if they had two points common, they would coincide and not intersect. Hence, two intersecting lines determine a point. 49. Axiom. --I straight line is the shortest line that can be drawn from one point to another. 50. Def. The distance between two points is the length of the straight line that joins tlieui. 51. A straight line determined by two points may be con- sidered as prolonged indefinitely. 52. If only the pai-t of the line between two fixed points is considered, this part is called a segment of the line. 53. For brevity, we say " the line AB," to designate a seg- ment of a line limited by the points A and B. 54. If a line is considered as extending from a fixed point, this point is called the origin of the line. * The general axioms on page 6 apply to all magnitudes. Special geometrical axioms will be given when required. THE STEAIGHT LINE. 9 55. If any point, C, is taken in a given straight line, AB, the two parts CA and CB are said to have opposite directions J ^-^ from the point C (Fig. 5). fig. b. Every straight line, as AB, may be considered as extending in either of two opposite directions, namely, fioni A towards B, which is expressed by AB, and read segment AB; and from B towai'ds A, which is expressed by BA, and read seg- ment BA. 56. If the magnitude of a given line is changed, it becomes longer or shorter. Thxis (Fig. 5), by prolonging ^C to 5 we add CB to AC, and AB = AC + CB. By diminishing AB to C, we subtract CB from AB, and AC = AB - CB. If a given line increases so that it is prolonged by its own magnitude several times in succession, the line is viultiplied, and the resulting line is called a multiple of the given line. Thus (Fig. 6),itAB=BC=^ ^ CD = DM, then AC = 2 AB, AB \ — i 1 1 T = ZAB,2ta&AE = iAB. Henoe, Fig. 6. Lines of given length may he added and subtracted ; they may also he multiplied hy a numher. THE PLANE ANGLE. 57. The opening between two straight lines drawn from the same point is called a plane angle. The two lines, ED and EF, are called the sides, and E, the point of meeting, is called the vertex of the angle. The size of an angle depends upon the extent ^<*- ''■ of opening of its sides, and not upon the length of its sides. 10 BOOK I. PLANE GEOMETRY. 58. If there is but one angle at a given vertex, the angle is designated by a capital letter placed at the vertex, and is read by simply naming the letter. If two or more angles have the same vertex, each angle is designated by three letters, and is read by naming the three letters, the one at the vertex between the "others. Thus, DAC (Tig. 8) is the ^ ^ ^^ angle formed- by the sides AD and AC. An angle is often designated by placing a small italic letter between the sides and near the vertex, as in Fig. 9. Fig. 9. 59. Postulate of Superposition. Any figure may be moved from one place to another without altering ics size or shape. 60. The test of equality of two geometrical magnitudes is that they may be made to coincide throughout their whole extent. Thus, Two straight lines are equal, if they can be placed one upon the other so that the points at their extremities coincide. Two angles are equal, if they can be placed one upon the other so that their vertices coincide and their sides coincide, each with each. 61. A line or plane that divides a geometric magnitude into Uvn equal parts is called the bisector of the magnitude. If the angles BAD and CAD (Fig. 8) are equal, AD biseets the angle BAC. 62. Two angles are called adjacent angles when they have the same vertex and a com- mon side between them; as the angles BOD and AOD (Fig. 10). THE PLANE ANGLE. 11 63. When one straight line meets ' another straight line and makes the adjacent angles equal, each of these angles is called a right angle; as angles DC A and DCB (Tig. 11). ^ q g Fig. 11. 64. A perpendicular to a straight line is a straight line that makes a right angle with it. Thus, if the angle DC A (Fig. 11) is a right angle, DC is perpendicular to AB, and AB is perpendicular to DC. 65. The point (as C, Fig. 11) where a perpendicular meets another line is called the foot of the perpendicular. 66. When the sides of an angle extend in opposite directions, so as to be in the same straight line, the angle is called a straight angle. A< .- >S Fig. 12. Thus, the angle formed at C (Fig. 12) with its sides CA and CB extend- ing in opposite directions from C is a straight angle. 67. Cob. A right angle is half a straight angle. 68. An angle less than a right angle is called an acute angle; as, angle A (Fig. 13). Fig. 13. 69. An angle greater than a right angle and ^ less than a straight angle is called an obtuse f' nI^^,^ angle; as, angle AOD (Fig. 14). "'""no u. 70. An angle greater than a straight angle and less than two straight angles is called a reflex angle; as, angle DO A, indicated by the dotted line (Fig. 14). 12 BOOK I. PLANE GEOMETRY. •71. Angles that ai-e neither right nor straight angles are called oblique angles ; and intersecting lines that aie not per- pendicular to each other aie called oblique lines. EXTENSION OF THE MEANING OF ANGLES. 72. Suppose the straight line OC (Fig. 15) to move in the plane of the paper from coincidence with OA, about the point as a pivot, to the position OC; then the line OG describes or generates the angle AOC, and the magnitude of the angle AOC depends upon the amount of rotation of the line from the position OA to the position OC. If the rotating line moves from the posi- tion OA to the position 07?, perpendicular to OA, it generates the right angle AOB; if it moves to the position OD, it generates the obtuse angle AOD; if it moves to the position OA', it generates the straight angle AOA' ; if it moves to the position 07?', it generates the reflex angle AOB', indicated by the dotted line ; and if it moves to the position OA again, it generates two straight angles. Hence, 73. The angular magnitude about a point in a plane is equal to two straight angles, or four right angles ; and the angular magnitude about a point on one side of a straight line drawn through the point is equal to a straight angle, or tivo right angles. 74. The who''e angular magnitude about a point in a plane is called a perigon ; and two angles whose sum is a perigon are called conjugate angles. Note. This extension of the meaning of angles is necessary in the applications of Geometry, as in Trigonometry, Mechanics, etc. THE PLANE ANGLE. 13 75. When two angles have the same vertex, and the sides of the one ai'e prolongations of the sides of the other, they are called vertical angles; as, angles a and b, c and d (Fig. 16). 77. Two angles are called supplementary when their sum is equal to a straight angle; and each is called the supplement of the other ; as, angles DOB and BOA (Fig. 18). ■(T FIQ. 16. 76. Two angles are called complementary when ^ their sum is^ equal to a right angle ; and each is called the com2}lement of the other; as angles DOB and DOC (Fig. 17). O B Pig. 17. O Fig. 18. UNIT OF ANGLES. 78. By adopting a suitable unit for measuring angles we are able to express the magnitudes of angles by numbers. If we suppose OC (Fig. 15) to turn about from coinci- dence with OA until it makes one three hundred sixtieth of a revolution, it generates an angle at 0, which is taken as the unit for measuring angles. This unit is called a degree. The degree is subdivided into sixty equal pai'ts, called ■minutes, and the minute into sixty equal parts, called seconds. Degrees, minutes, and seconds are denoted by symbols. Thus, 6 degrees 13 minutes 12 seconds is written 6° 13' 12". A right angle is generated when OC has made one fourth of a revolution and contains 90° ; a straight angle, when OC has made half of a revolution and contains 180° ; and a perigon, when OC has made a complete revolution and contains 360°. KoTte. The natural angular unit is one complete revolution. But this unit would require us to express the values of most angles by frac- tions. The advantage of using the degree as the unit consists in its con- venient size, and In the fact that 360 is divisible by so many different integral numbers. 14 BOOK I. PLAira; GEOMETKY. D B Fig. 19. 79. By the method of superposition we are able to compare magnitudes of the same kind. Suppose we have two angles, ABCaja&I)EF(Fig.lOi). Let the side MB be placed on the side BA, so that the vertex E shall fall on B ; then, if the side EF falls on ^ C, the angle DEF equals the angle ABC; if the side EF falls between BC and BA in the position shown by the dotted line BG, the angle DEF is less than the angle ABC; but if the side EF falls in the position shown by the dotted line BH, the angle DEF is greater than the angle ABC. G F / / / / ■M FlO. 20. B A Fia 21. 80. If we have the angles ABC and DEF (Fig. 20), and place the vertex E on B and the side ED on BC, so that the angle DEF takes the position CBIT, the angles DEF and ABC will together be equal to the angle ABIT. If the vertex E is placed on B, and the side ED on BA, so that the angle DEF takes the position ABF, the angle FBC will be the difference between the angles ABC and DEF. If an angle is increased by its own magnitude two or more times in succession, the angle is multiplied by a number. Thus, if the angles ABM, MBC, CBP, PBD (Fig. 21) are all equal, the angle ABD is 4 times the angle ABM. Therefore, Angles may he added and subtracted ; they may also be mul- tiplied by a number. PERPENDICULAR AND OBLIQUE LINES. 15 PERPENDICULAR AND OBLIQUE LINES. Pkopositiok I. Theoeem. 81. All straight angles are equal. A '- B D F Let the angles ACB and DEF be any two straight angles. To prove that /LACB = Z. DEF. Proof. Place the Z. ACB on the Z DEF, so that the vertex C shall fall on the vertex E, and the side CB on the side EF. Then CA will fall on ED, § 47 (peeause ACB and DEF are straight lines). .■ . A ACB =^ A DEF. §60 Q. E. D. 82. CoK. 1. Ml right angles are equal. Ax. 7 83. CoE. 2. At a given 'point in a given line there can he but one perpendicular to the line. For, if there could be two Js, we should have / rt. A of different magnitudes ; but this is impos- / sible, § 82. ^ 84. CoE. 3. The complements of the same angle or of equal angles are equal. Ax. 3 85. Cor. 4. The supplements of the same angle or of equal angles are equal. Ax. 3 Note. The beginner must not forget that in Plane Geometry all the points of a figure are in the same plane. Without this restriction in Cor. 2, an indefinite number of perpendiculars can be erected at a given point in a given line. 16 BOOK I. PLANE GEOMETRY. Proposition II. Theokem. 86. If two adjacent angles have their exterior sides in a straight line, these angles are supplementary. Let the exterior sides OA and OB of the adjacent angles ADD and BOD be in the straight line AB. T(i prove that A AOD and BOD are supplementary. Proof. AOB is a straight Une. Hyp. .-.Z^O^is ast. Z. §66 But ZAOB+Z BOD = the st. Z AOB. Ax. 9 -■. the A AOD and BOD ai-e supplementary. § 77 Q.E.D. 87. Def. Adjacent angles that axe supplements of each other are called SKjijjhmentary-adjacent angles. Since the angular magnitude about a point is neither in- creased nor diminished by the number of lines which radiate from the point, it follows that, 88. Cor. 1. The sum of all the angles about a point in a plane is equal to a perigon, or two straight angles. 89. CoE. 2. The sum of all the angles about a point in a plane., on the same side of a straight line passim/ through the point, is equal to a straight angle, or two right angles. PERPENDICULAR AND OBLIQUE LINES. 17 Proposition III. Theorem. 90. Conversely: If two adjacent angles are supple- mentary, their exterior sides are in the same straight line. —-F B Let the adjacent angles OCA and OCB be supplementary. To prove that AC and CB are in the same stravjht line. Proof. Suppose C-^'to be in the same line with AC- Then ^ OC^ and OCi?" are supplementary, §86 {if two adjacent angles have their exterior sides in u, straight line, these angles are supplementary). But A OCA and OCB are supplementary. Hyp. .'. A OCF and OCB have the same supplement. .■.AOCF=Z.OCB. §85 .-. CB and CF coincide. § 60 .'.AC and CB are in the same straight line. q.e.d. Since Propositions II. and III. are true, their opposites are true. Hence, §.33 91. Cor. 1. If the exterior sides of two adjacent angles are not in a straight line, these angles are not supplementary. 92. Cor. 2. J^ two adjacent angles are not supplementary/, their exterior sides are not in the same straight line. 18 BOOK I. PLANE GEOMETRY. Proposition IV. Theorem. 93. If one straight line intersects another straight line, the vertical angles are equal. -p Let the lines OP and AB intersect at C. To prove that AOCB = /.A CP. Proof. A OCA and AOCB are supplementary. § 86 Z OCA and Z.ACP are supplementary, § 86 (if two ac^acent angles have their exterior sides in a straight line, these angles are supplemeniary). .'.A OCB and ACP have the same supplement. .■.AOCB = AACP. §85 In like manner, AACO = AP CB. Q.E.D. 94. Cor. If one of the four angles formed by the inter- section of two straight lines is a right angle, the other three angles are right angles. Ex. 1. Find the complement and the supplement of an angle of 49°. Ex. 2. Find the numher of degrees in an angle if it is double its com- plement ; if it is one fourth of its complement. Ex. 3. Find the numher of degrees in an angle if it is double its sup- plement ; if it is one third of its supplement. PERPENDICULAR AND OBLIQUE LINES. 19 Peopositioit v. Theorem. 95. Two straight lines drawn from a point in a per- pendicular to a given line, cutting off on the given line equal segments from the foot of the perpendicular, are equal and make equal angles with the perpendicular. Let CF be a perpendicular to the line AB, and CE and CK two straight lines cutting off on AB equal segments FE and FK from F. To prove that CE = CK; and Z FCE = Z FCK. Proof. Pold over CFA, on CF as an axis, until it falls on the plane at the right of CF. FA will fall along FB, (since /. CFA = Z CFB, each being a rt. /., hy hyp.). Point ^will fall on point K, (since FE = FK, by hyp.). .■.CF=CK, §60 Ifheir extremities being the same points); and Z FCE = Z FCK, § 60 (since th&,r vertices coincide, and their sides coincide, each with each). Q. E. D. Ex. 4. Find the number of degrees in the angle included by the hands of a clock at 1 o'clock. 3 o'clock. 6 o'clock. 6 o'clock. 20 BOOK I. PLANE GEOMETRY. Peoposition VI. Theorem. 96. Only one perpendicular can he drawn to a given line from a given external point. Let AB be the given line, P the given external point, PC a perpen- dicular to AB from P, and PD any other line from P to AB. To prove that PD is not J_ to AB. Proof. Produce PC to P', making GP' equal to PC. Draw DP'. By construction, PCP' is a straight line. . PDP' is not a straight line, § 46 {only one straight line can be drawn from one point to another). Hence, /.PDP' is not a straight angle. Since PC is ± to Z»C, and PC = CP', AC is J_ to PP' at its middle poiiit. .■ /-PDC = Z.P'DC, § 95 {two straight lines from a point in a ± to a line, cutting off on the line equd segments from the foot of the X, make equal A with the _L). Since Z PDP' is not a straight angle, Z PDC, the half of A PDF', is not a right angle. .-. PP< is not ± to ^£. „„„ O.E.D. PERPENDICULAR AND OBLIQUE LINES. 21 Proposition VII. Theorem. 97. The perpendicular is the shortest line that can he draion to a straight line from an external point. Let AB be the given straight line, P the given point, PC the per- pendicular, and PD any other line drawn from P to AB. To prove that PC < PD. Proof. Produce PC to P', making CP' = PC. Draw DP'^ Then PD = DP', § 95 (two straight lines drawn from a point in a Xto a line, cutting off on the line equal segments from the foot of the X, are equal). . . PD +DP' = 2 PD, and PC + CP' = 2 PC. Const. But PC + CP' OA + OB. Proof. Produce AO to meet the line CB at H. Then CA+CE>OA+ OE, and BE + OE > OB, § 49 (a straight line is the shortest line from one point to another). Add these inequalities, and we have CA+GE+BE + OE>OA+OE + OB. Ax. 4 Substitute for CE +BE its equal CB, then CA + CB + OE>OA + OE+OB. Take away OE from each side of the inequality. CA + CB>OA + OB. A.X. 5 Q. E. D. PERPENDICTILAR AND OBLIQUE LINES. 23 Peoposition IX. Theorem. 101. Of two straight lines drawn from the same point in a perpendicular to a given line, cutting off on the line unequal segments from the foot of the perpendicu- lar, the more remote is the greater. Let OC be perpendicular to AB, OG and OE two straight lines to AB, and CE greater than C6. To prove that OE>OG. Proof. Take CF equal to CG, and draw OF. Then 0F= OG, § 95 {two straight lines drawn from a point in a A- to a line, cutting off on the line equal segments from the foot of the ±, are equal). Produce OC to D, making CD = OC. Draw FD and FB. Then OF = FJD, and OF = FD. § 95 But OF +FD > OF +FD, § 100 .■.2 0F>2 0F, OF>OF, a.nd OF >0G. q.e.d. 102. Cor. Onli/ two equal straight lines can he drawn from a point to a straight line; and of two unequal lines, the greater cuts off on the line the greater segment from the foot of the perpendicular. 24 BOOK I. PLANE GEOMETRY. PARALLEL LINES. 103. Dep. Two parallel lines are lines that lie in the same plane and cannot meet however far they are produced. Peoposition X. Theorem. 104. Tloo straight lines in the same plane perpen- dicular to the same straight line are parallel. -B Let AB and CD be perpendicular to AC. To prove that AB and CD are parallel. Proof. If AB and CD are not parallel, they will meet if sufficiently prolonged, and we shall have two perpendicular lines from their point of meeting to the same straight line ; but this is impossible, ' § 96 (only one perpendicular can be drawn to a given line from a given external point). .-. AB and CD are parallel. q.e.d. 105. Axiom. Through a f/iren point only one straight line can be drawn parallel to a given straight line. 106. Cor. Two straight lines in the same plane parallel to a third straight line are parallel to each other. For if they could meet, we should have two straight lines from the point of meeting parallel to a straight line ■ but this is impossible. § ;[05 PARALLEL LINES. 25 Proposition XI. Theorem. 107. If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other also. H E K N Let AB and EF be two parallel lines, and let HK be perpendicular to AB, and cut EF at C. To prove that HK is ±to HF. Proof. Suppose MN drawn thiough C _L to HK. Then MN is II to AB. But But HF is II to AB. EF coincides with MN. MN is -L to HK. .-. EF is ± to HK, HK is J_ to EF. § 104 Hyp. §106 Const. that is, nn- is _i_ to jiij:. q.e.d. 108. Def. a straight line that cuts two or more straight lines is called a transversal of those lines. 109. If the transversal EF cuts AB and CD, the angles a, d, g,f are called interior angles ; b, c, h, e are called exterior angles. The angles d and /, and a and g, are called alternate-interior angles ; the angles b and h, and c and e, are called alternate-exterior angles. The angles b and /, c and g, e and a, h and d, are called exterior-interior angles. 26 BOOK I. PLANE GEOMETRY. Proposition XII. Theorem. 110. If two parallel lines are cut by a transversal, the alternate-interior angles are equal. E —4 -^ F Let EF and GH be two parallel lines cut by the transversal BC. To prove that AEBC = Z.BCH. Proof. Thiough 0, the middle point of BC, suppose AD drawn _L to GH. Then AD is likewise ± to EF, § 107 (a straight line ± to one of two lis is J. to the other), that is, CD and BA are both _L to AD. Apply the figui'e COD to the figuie BOA, so that OD shall fall along OA. Then OC will fall along OB, § 93 (since Z COD = Z BOA, being vertical A); and C will fall on B, (since OC = OB, by construction). Then the ± CD will fall along the ± BA, § 96 (only one L can be drawn to a given line from a given ezternal point). .-. ZOCD coincides with Z OB A, and is equal to it, § 60 (two angles are equal, if their vertices coincide and their sides coincide each with each). 0. E. D. PARALLEL LINES. 27 Proposition XIII. Theorem. 111. Conversely : When tivo straight lines in the same plane are cut hy a transversal, if the alternqi&dMerior angles are equal, the two straight lines are parallel. A """"--^^^-_ B Let EF cut the straight lines AB and CD in the points H and K, and let the angles AHK and HKD be equal. To prove that AB is II to CD. Proof. Suppose MJSf drawn through If II to CD. Then Z MJIK = Z HKD, § 110 {being alt. -int. ^ 0/ II lines). But Z ASK = Z HKD. Hyp. .-. Z MHK = Z AHK. Ax. 1 .•. MN and AB coincide. § 60 But MN is II to CD. Const. .■. AB, which coincides with MN, is II to CD. Q. E. D. Ex. 5. Find the complement arid the supplement of an angle that con- tains 37° 53' 49". Ex. 6. If the complement of an angle is one third of its supplement, how many degrees does the angle contain ? 28 BOOK I. PLANE GEOMETRY. Proposition XIV. Theorem. 112. If tioo parallel lines are cut by a transversal, the exterior-interior angles are equal. E F Let AB and CD be two parallel lines cut by the transversal EF, in the points H and K. To prove that Z UHB = Z HKD. Proof. Z EHB = Z AHK, § 93 {being vertical A). Z AHK = Z HKD, § 110 (being alt. -int. A of II lines). .-. Z KHB = Z HKD. Ax. 1 In like mauner Z HHA = Z HKC. q. e. d 113. Cor. The alternate-exterior angles EHB and CKF, and also AHE and DKF, are equal. Proposition XV. Theorem. 114. Conversely : Whe7i two straight lines in a plane are cut by a transversal, if the exterior-interior angles are equal, these tioo straight lines are p)araTlel. (Proof similai to that in § 111.) PARALLEL LINES. 29 Proposition XVI. Theorem. 115. If two parallel lines are cut by a transversal, the tioo interior angles on the same side of the transversal are supplementary. E ~D Let AB and CD be two parallel lines cut by the transversal EF in the points H and K. To prove that A BHK and HKD are supplementary. Proof. Z EHB + Z BHK = a st. Z, § 86 (being sup.-adj. A). But Z EBB = Z HKD, § 112 [being ext.-int. AofW lines). .-. Z BHK + Z HKD = a st. Z. .'.A BHK and HKD are supplementary. § 77 Q.E.D. Proposition XVII. Theorem. 116. Conversely : When tivo straight lines in a j^lane are cut by a transversal, if tioo interior angles on the same side of the transversal are supplementary, the tioo straight lines are parallel. (Proof similar to that in § 111.) 30 BOOK I. PLANE GEOMETRY. TRIANGLES. 117. A triangle is a portion of a plane bounded by tkree straight lines ; as, ABC (Fig. 1). The bounding lines are called the sides of the triangle, and their sum is called its perimeter ; the angles included by the sides are called the angles of the triangle, and the vertices of these angles, the vertices of the triangle. 118. Adjacent angles of a rectilinear figui-e are two angles that have one side of the figure common; as, angles A and B (Fig. 2). fig. 2. 119. An exterior angle of a triangle is an angle included by one side and another side produced; as, ACD (Fig. 2). The interior angle ACB is adjacent to the exterior angle; the inte- rior angles, A and B,'w& called opposite interior angles. Scalene. Isosceles. Equilateral. 120. A triangle is called a scalene triangle when no two of its sides are equal ; an isosceles triangle, when two of its sides are equal ; an equilateral triangle, when its three sides are equal. Bight. Obtuse. Acute. Equiangular, i A TRIANGLES. 31 121. A triangle is called a right triangle, when one of its angles is a right angle ; an obtuse triangle, when one of its angles is an obtuse angle; an acute triangle, when all three of its angles are acute angles ; an equiangular triangle, when its three angles are equal. 122. In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two sides the legs. 123. The side on which a triangle is supposed to stand is called the base of the triangle. In the isosceles triangle, the equal sides are called the legs, and the other side, the base ; in other triangles, any one of the sides may be taken as the base. 124. The angle opposite the base of a triangle is called the vertical angle, and its vertex, the vertex of the triangle. 125. The altitude of a triangle is the perpendicular from the vertex to the base, or to the base produced ; as, AD (Fig. 1). 126. The three perpendiculars from the vertices of a triangle to the opposite sides (produced if necessary) are called the altitudes of the triangle ; the three bisectors of the angles are called the bisectors of the triangle; and the three lines from the vertices to the middle points of the opposite sides are called the medians of the triangle. 127. If two triangles have the angles of the one equal, respec- tively, to the angles of the other, the equal angles are called homologous angles, and the sides opposite the equal angles are called homologous sides. 128. Two triangles are equal in all respects if they can be made to coincide (§ 60). The homologous sides of equal tri- angles are equal, and the homologous angles are equal. 82 BOOK I. PLANE GEOMETRT, Proposition XVIII. Theorem. 129. Tlie sum of the three angles of a triangle is equal to two right angles. B p ■^ c" J^ Let A, B, and BCA be the angles of the triangle ABC. To prove that Z A + Z B + Z BCA = £ rt. A. Proof. Suppose CJS drawn II to AB, and prolong AC to F. Then ZECF +ZECB +ZBCA = 2T:t. A, §89 (the sum of all the A about a point on the same side of a straight liiie passing through the point is equal to Srt. A). But ZA = Z ECF, § 112 ifieing ext.-int. A of the II lines AB and CM), and ZB = Z BCE, § 110 {being alt.-int. A of the II lines AB and CE). Put for the A ECF and BCE their equals, the A A and B. Then ZA+ZB +ZBCA = 2 rt. A. q.e.d. 130. Cor. 1. The sum of two angles of a triangle is less than two right angles. 131. Cor. 2. If the sum of two angles of a triangle is taken from two right angles, the remainder is equal to the third angle. 132. Cor. 3. If two triangles have two angles of the one equal to two angles of the other, the third angles are equal. TRIANGLES. 33 133. Cor. 4. If two right triangles have an acute angle of the one equal to an acute angle of the other, the other acute angles are equal. 134. Cob. 5. In a triangle there can he hut one right angle, or one ohtuse angle. 135. Cob. 6. In a right triangle the two acute angles are together equal to one right angle, or 90°. 136. Cob. 7. In an equiangular triangle, each angle is one third of two right angles, or 60°. 4/137. Cob. 8. An exterior angle of a triangle is equal to the sum of the two opposite interior angles, and therefore greater than either of them. Proposition XIX. Theobem. 138. Tlie sum of two sides of a triangle is greater than the third side, and their difference is less than the third side. B A O In the triangle ABC, let AC be the longest side. To prove that AB + BC >AC, and AC ~ BC < AB. Proof. AB +BOAC, § 49 (a straight line is the shortest line from one point to another). Take away BC from both sides. Then AB>AC-BC, Ax. 5 or AC- BC AC. Proof. Now AB = AC, or AC. But AB is not equal to AC; for then the Z C would be equal to the Z.B, § 146 (being A opposite equal sides). And AB is not less than A C ; for then the Z C would be less than the Z 5. § 162 Both these conclusions are contrary to the hypothesis that the Z C is greater than the Z B. Hence, AB cannot be equal to ^C or less than AC. .■.AB>AC. Q.E.D. Ex. 11. If the vertical angle of an isosceles triangle is equal to 30°, find the exterior angle included by a side and the base produced. Ex. 12. If the vertical angle of an isosceles triangle is equal to 36°, find the angle included by the bisectors of the base angles. 42 BOOK I. PLANE GEOMETRY. Proposition XXVIII. Theorem. 154. If two triangles have two sides of the one equal, respectively, to two sides of the other, hut the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second. In the triangles ABC and ABE, let AB be equal to AB, BC to BE ; but let the angle ABC be greater than the angle ABE. To prove that AOAE. Proof. Place the A so that AB of the one shall fall on AB of the other, and BE within the A ABC. Suppose BF drawn to bisect the Z MB C, and draw EF. The A FBF and CBF are equal. § 143 For BF = BF, Iden. BF = BC, Hyp. and Z FBF = Z CBF. Const. .■.FF=FC. §128 Now AF + FE > AE. § 133 .-. AF + FC > AE. ■•^G>AE. ^^^ TRIANGLES. 43 Proposition XXIX. Theokbm. 155. Conversely : If two sides of a triangle are equal, respectively, to two sides of another, hut the third side of the first triangle is greater than the third side of the second, then the angle opposite the third side of (he first triangle is greater than the angle opposite the third side of the second. In the triangles ABC and DEF, let AB be equal to DE, AC to DF ; but let BC be greater than EF. To prove that the ^ A is greater than the Z. D. Proof. Now the Z-A'vs, equal to the Z. D, or less than the Z.D, or greater than the Z. D. But the Z A is not equal to the Z D ; for then the A ABC would be equal to the A DUF, § 143 {frnving two sides and the included Z of the one equal, respectively, to two sides and the included Z of the other), and BC would be equal to UF. And the ZA is not less than the Z D, for then BC would be less than FF. § 154 Both these conclusions are contrary to the hypothesis that ^C is greater than FF. Since the ZAis not equal to the ZD or less than the Z D, the Z ^ is greater than the Z D. q.e.d. 44 BOOK I. PLANE GEOMETEY. LOCI OF POINTS. 156. If it is required to find a point which shall fulfil a sinffle geometric condition, the point may have an unlimited number of positions. If, however, all the points are in the same plane, the required point will be confined to a particular line, or group of lines. A point in a plane at a given distance from a fixed straight line of indefinite length in that plane, is evidently in one of two straight lines, so drawn as to be everywhere at the given distance from the fixed line, one on one side of the fixed line, and the other on the other side of it. A point in a plane equidistant from two parallel lines in that plane is evidently in a straight line drawn between the two given parallel lines and everywhere equidistant from them. 157. All points in a plane that satisfy a single geometrical condition lie, in general, in a line or group of lines ; and this line or group of lines is called the locus of the points that satisfy the given condition. 158. To prove completely that a certain line is the locus of points that fulfil a given condition, it is necessary to prove 1. Any point in the line satisfies the given condition; and any poiiit not in the line does not satisfy the given con- dition. Or, to prove 2. Any point that satisfies the given condition lies in the line; and any point in the line satisfies the given condition. Note. The word locus (pronounced lolius) is a Latin word that signi- fies pface. The plural of locus is loci (pronounced lo'si). 159. Def. a line which bisects a given line and is perpen- dicular to it is called the perpendicular bisector of the line. LOCI OF POINTS. 45 Proposition XXX. Thboeem. 160. The perpendicular bisector of a given line is the locus of points equidistant from the extremities of the line. A.^ n. D '-^B Let PR be the perpendicular bisector of the line AB, any point in PR, and C any point not in PR. Draw OA and OB, CA and CB. To prove OA and OB equal, CA and CB unequal. Proof. 1. A OFA = A OPB, § 144 for PA = PB by hypothesis, and OP is common, (f,wo right A are equal if their legs are equal, each to each). .-. OA = OB. % 128 2. Since C is not in the ±, CA or CB will cut the J.. Let CA cut the ± at B, and draw BB. Then, by the first part of the proof DA = DB. But CB and U coincides with this parallel and is II to BC. Also, since EF drawn II to AB bisects AC, it bisects ^C, by § 188 ; that is, BF=FC = \BC. But BDEF is a O by § 166, and therefore DE = BF=iBC. 190. CoR. 3. The median of a trapezoid is parallel to the bases, and is equal to half the sum of the bases. B D a /■~~ / \ \ / ^^ \ F.I \ \n 1 F ~~v ^--\ Draw the diagonal DB. In the A ADB join E, the middle point of AD, to F, the middle point of DB. Then, by § 189, EF is II to AB and = ^AB. In the ADBC join F to G, the middle point of BC. Then FG is II to DC and = IDC. AB and FG, being II to DC, are II to each other. But only one line can be drawn through F II to AB (§ 105). Therefore FG is the prolongation of EF. Hence, EFG is paxallel to AB and DC, and equal to \ {AB + DC). 56 BOOK I. PLANE GEOMETRY. POLYGONS IN GENERAL. 191. A polygon is a portion of a plane bounded by straight lines. The bounding lines are the sides, and their sum, the perim- eter of the polygon. The angles included by the adjacent sides are the angles of the polygon, and the vertices of these angles are the vertices of the polygon. The number of "sides of a polygon is evidently equal to the number of its angles. 192. A diagonal of a polygon is a line joining the vertices of two angles not adjacent; as, AC (Fig. 1)- B E Fig. 1. Fig. 3. 193. An equilateral polygon is a polygon which has all its. sides equal. 194. An equiangular polygon is a polygon which has all its angles equal. 195. A convex polygon is a polygon of which no side, when produced, will enter the polygon. 196. A concave polygon is a polygon of which two or more sides, if produced, will enter the polygon. 197. Each angle of a convex polygon (Fig. 2) is called a salient angle, and is less than a straight angle. 198. The angle HDF of the concave polygon (Fig. 3) is called a re-entrant angle, and is greater than a straight angle. When the term polygon is used, a convex polygon is meant. POLYGONS. 57 199. Two polygons are eqxial when they can be divided by diagonals into the same number of triangles, equal each to each, and similarly placed ; for if the polygons are applied to each other, the corresponding triangles will coincide, and hence the polygons will coincide and be equal. 200. Two polygons ai-e mutually equiangular, if the angles of the one are equal to the angles of the other, each to each, when taken in the same order. Figs. 1 and 2. 201. The equal angles in mutually equiangular polygons are called homologous angles ; and the sides which ai'e included by homologous angles are called homologous sides. 202. Two polygons are mutually equilateral, if the sides of the one are equal 'to the sides of the other, each to each, when taken in the same order. Figs. 1 and 2. Fig. 4. Fig. B. Fig. 6. Fia. 7. 203. Two polygons may be mutually equiangular without being mutually equilateral ; as. Figs. 4 and 5. And, except in the case of triangles, two polygons may be mutually equilateral without being mutually equiangular ; as, Figs. 6 and 7. If two polygons are mutually equilateral and mutually equi- angular, they are equal, for they can be made to coincide. 204. A polygon of three sides is called a triangle ; one of four sides, a quadrilateral ; one of five sides, 2^ pentagon ; one of six sides, a hexagon ; one of seven sides, a heptagon ; one of eight sides, an octagon ; one of ten sides, a decagon ; one of twelve sides, a dodecagon. 68 BOOK I. PLANE GEOMETRY. Proposition XXXIX. Theoeem. 205. The sum of the interior angles of a polygmi is equal to two right angles, taken as viani^i times less two as the figure has sides. Let the figure ABCDEF be a polygon, having n sides. To prove that AA + Z.B + AC, etc. = (re - 2) 2 rt. A. Proof. Prom A diaw the diagonals AC, AD, and AE. The siun of the A of the A is equal to the sum of the A of the polygon. Xow, there are Qi — 2) A, and the sum of the A of each A = 2 rt. ^. § 129 . . the sum of the A of the A, that is, the sum of the A of the polygon is equal to (re — 2)2 rt. A. q.e.d. 206, Cor. The sum of the angles of a quadrilateral equals 4 right angles ; and if the angles are all equal, each is a right angle. In general, each angle of an equiangular polygon of . ^ . 2 (« - 2) . , n sides is equal to right angles. Ex. 17. How many diagonals can be drawn in a polygon of n sides ? POLYGONS. 59 Proposition XL. Theorem. 207. The exterior angles of a polygon, made hy pro- ducing each of its sides in succession, are together equal to four right angles. Let the figure ABODE be a polygon, having its sides produced in succession. To prove the sum of the ext. A = irt A. Proof. Denote the int. A of the polygon by A, B, C, D, E, and the corresponding ext. A by a, b, c, d, e. AA + Za = 2it.A, § 89 and AB + Zb =2Tt. A, {being sup.-adj. A). In like manner each pair of adj. ^ = 2 rt. A. .'. the sum of the interior and exterior A of a, polygon of n sides is equal to 2n rt. A. But the sum of the interior A=(n — 2)2 rt. A, § 206 = 2 w rt. ^ — 4 rt. A. .'. the sum of the exterior A = i rt. A. q.e.d. Ex. 18. How many sides has a polygon if the sum of its interior A is twice the sum of its exterior A f tea times the sum of its exterior A ? 60 BOOK I. PLANE GEOMETRY. SYMMETRY. 208. Two points are said to be S3nnmetrical with respect to a third point, called the centre of symmetry, if this third point bisects the straight line which joins them. X— Two points are said to be symmetrical with respect to a straight line, called the axis of symmetry, if this straight line bisects at right angles the straight line which joins them. Thus, P and P' are symmetrical with respect to O as a centre, and XX' as an axis, if bisects the line PP', and if XX' bisects PP' at right angles. 209. A figure is symmetrical with respect to a point as a centre of symmetry, if the point bisects every straight line drawn through it and terminated by the boundary of the figure. 210. A figure is symmetrical with respect to a line as an axis of symmetry if one of the parts of the figure coincides, point for point, with the other part when it is folded over on that line as an axis. 211. Two figures are said to be symmet- rical with respect to an axis if every point of one has a corresponding symmetrical point in the other. Thus, if every point in the figure A'B'C has a. symmetrical point in ABC, with respect to XX' as an axis, the figure A'B'C is symmetrical to ABC with respect to XX' as an axis. X~ Ji SYMMETRY. 61 Proposition XLI. Theorem. 212. A quadrilateral which has two adjacent sides equal, and the other two sides equal, is symmetrical loith respect to the diagonal joining the vertices of the angles formed by the equal sides, and the diagonals are perpen- dicular to each other. Let ABCD be a quadrilateral, having AB equal to AD, and CB equal to CD, and having the diagonals AC and BD. To prove that the diagonal AC is an axis of symmetry, and that it is X to the diagonal BD. Proof. In the A ABC and ADC, AB = AD, and BC = DC, and AC = AC. .-.A ABC = A ADC. Hyp. Iden. §150 .-. Z BAC = Z DAC, and Z BCA = Z DCA. Hence, if ABC is turned on ^C as an axis until it falls on ADC, AB will fall upon AD, CB on CD, and OB on OD. .-.the A ABC will coincide with the A ADC. .■. ^C is an axis of symmetry (§ 210) and is ± to BD. § 208 Q.E.D. 62 BOOK I. PLANE GEOMETRY. Pkoposition XLII. Theorem. 213. If a figure is symmetrical with respect to two axes perpendicular to each other, it is symmetrical with respect to their intersection as a centre. 1 Y B C r^ M .?~^JV A ^ //'O r» H K T G 1 JP Y Let the figure ABCDEFGH be symmetrical with respect to the two perpendicular axes XX', YY', which intersect at 0. To prove that is the centre of symmetry of the figure. Proof. Let N be any point in the perimeter. Suppose NMI drawn ± to YT, IKL ± to XJC. Then §104 NI is II to XX' and IL is II to YT. Draw LO, ON, and KM. Now KI = KL, (the figure being symmetrical with respect to XX'). But KI = OM. .-. KL = OM, and KLOM is a O. .'. ZO is equal and parallel to KM. In like manner ON is equal and parallel to KM. .-.LON is a straight line. § 105 .-.0 bisects LN, any straight line and therefore every straight line drawn through and terminated by the perimeter. -•. is the centre of symmetry of the figure. o b d §208 §180 §183 §183 REVIEW QUESTIONS. 63 REVIEW QUESTIONS ON BOOK I. 1. What is the subject-matter of Geometry ? 2. What is a geometric magnitude ? 3. What is an ajdom ? a theorem ? a converse theorem ? an oppo- site tlieorem ? a contradictory theorem ? 4. Define a straight line ; a curved line ; a broken line ; a plane sur- face ; a curved surface. 5. How many points are necessary to determine a straight line ? 6. How many straight lines are necessary to determine a point 1 7. On what does the magnitude of an angle depend ? 8. Define a straight angle ; a right angle ; an oblique angle. 9. Define adjacent angles ; complementary angles ; supplementary angles ; conjugate angles. 10. Define parallel lines and give the axiom of parallels. 11. If two lines in the same plane are parallel and cut by a trans- versal, what pairs of angles are equal ? what pairs are supplementary ? 12. Define a right triangle ; an isosceles triangle ; a scalene triangle. 13. To how many right angles is the sum of the angles of a triangle equal ? the sum of the acute angles of a right triangle ? 14. To what angles is the exterior angle of a triangle equal ? 15. What is the test of equality of two geometric magnitudes ? 16. How does a reciprocal theorem differ from a converse theorem ? 17. State the three cases in which two triangles are equal. 18. State the cases in which two right triangles are equal. 19. What is meant by a locus of points ? 20. Where are the points located in a plane that are each equidistant from two given points 1 from two intersecting lines ? 21. Define a parallelogram ; a trapezoid'; an isosceles trapezoid. 22. Wlien is a figure symmetrical with respect to a centre ? 23. When is a figure symmetrical with respect to an axis ? 24. Must a triangle be equiangular if equilateral ? must a triangle be equilateral if equiangular ? 25. When are two polygons said to be mutually equiangular ? 26. When are two polygons said to be mutually equilateral ? 27. dan two polygons of more than three sides be mutually equiangular without being mutually equilateral ? mutually equilateral without being mutually equiangular ? 28. What line do two points each equidistant from the extremities of a given straight line determine ? 64 BOOK I. PLANE GEOMETRY. METHODS OF PROVING THEOREMS. 214. There are three general methods of proving theorems, the synthetic, the analytic, and the indirect methods. The synthetic method is the method employed in most of the theorems already given, and consists in putting together known truths in order to obtain a new truth. The analytic method is the reverse of the synthetic method. It asserts that the conclusion is true if another proposition is true, and so on step by step, until a known truth is reached. Thus, proposition A is true if proposition B is true, and B is true if C is true ; but C is true, hence A and B are true. If a known truth suggests the required proof, it is best to use the synthetic form at once. If no proof occurs to the mind, it is necessary to use the analytic method to discover the proof, and then the synthetic proof may be given. The indirect method, or the method of reductio ad absurdum, is illustrated on page 41. It consists in proving a theorem to be true by proving its contradictory to be false. 215. Generally auxiliary lines are required, as a line con- necting two points ; a line parallel to or perpendicular to a given line ; a line produced by its own length ; a line making with another line an angle equal to a given angle. Two lines are proved equal by proving them homologous sides of equal triangles ; or legs of an isosceles triangle; or opposite sides of a parallelogram. Two angles are proved equal by proving them alternate-interior angles or exterior-interior angles of parallel lines ; or homolo- gous angles of equal triangles ; or base angles of an isosceles tri- angle ; or opposite angles of a parallelogram. Two suggestions axe of special importance to the beginner: 1. Draw as accurate figures as possible. 2. Draw as general figures as possible. EXERCISES. 65 Prove by the analytic method: Ex. 19. A median of a triangle is less than half the sum of the two adja- cent sides. A To prove the median AB <\{AB + AC). Now AD <^(AB + AC), if 2AD = Z.D. (Why?) .. AABD = AA + /.D. Ex. 44. A line drawn from the vertex of the right angle of a right tri- angle to the middle point of the hypotenuse divides the triangle into two isosceles triangles. Ex. 45. If the equal sides of an isosceles triangle are produced through the vertex so that the external segments are equal, the extremities of these segments will be equally distant from the extremities of the base, respectively. A Ex. 46. If through any point in the bisector of an / P angle a line is drawn parallel to either of the sides of /"^-^P the angle, the triangle thus formed is isosceles. -c^—- Ex. 47. Through any point C in the line AB an intersecting line is drawn, and from any two points in this line equidistant from C perpen- diculars are dropped on AB or AB produced. Prove that these per- pendiculars are equal. Ex. 48. If the median drawn from the vertex of a triangle to the base is equal to half the base, the vertical angle is a right angle. Ex. 49. The lines joining the middle points of the sides of a triangle divide the triangle into four equal triangles. 70 BOOK I. PLANE GEOMETET. Ex. 50. The altitudes upon the legs of an isosceles triangle are equal. Rt. A BEC = rt. A GDB (§ 141). Ex. 51. If the altitudes upon two sides of a triangle are equal, the tri- angle is isosceles. Rt. A BEG = rt. A CDB (§ 151). A Ex. 52. The medians drawn to the legs of an isosceles triangle are equal. A BEC = A CBB (§ 143). Ex. 53. If the medians to two sides of a triangle are equal, the triangle is isosceles. BO = CO, and OE = OD (Ex. 27). Z BOB = Z COD. .-. A BOE = A GOB (§ 143). Ex. 54. The bisectors of the hase angles of an isosceles triangle are ^'^"^^" A BEC = A CDB (§ 139). Ex. 55. Opposite Theorem. If a triangle is not isosceles, the bisectors of the base angles are not equal. Let Z ABC be greater than ZACB; then KC > KB. (Why ?) Now CD > BE, if KD is greater than or equal to KE. But suppose KD < KE. Lay ofi KH = KD and KG = KB, join SG, and draw 6F II to BE. A KDB = A KHG. (Why ?) .-. Z KHG = Z KDB. (Why ?) .-.Z-BTBC is greater than Z^ffG. (Why?) .. GF > HE. (Why?) Z GFC is greater than Z FCG {i ACB). .: CG > GF, and > HE. . . KC - KG > KE - KH, or KC + KD > KB + KE, or CD > BE. Ex. 56. State the converse theorem of Ex. 54. Is the converse theo- rem true ? Ex. 57. The perpendiculars dropped from the middle ^ point of the base upon the legs of an isosceles triangle are equal. A BED = A CFD (§ 141). A. Ex. 58. State and prove the converse. B D C A BED = A CFD (§ 151). EXERCISES. 71 Ex. 59. The difference of the distances from any point in the base produced of an isosceles triangle to the equal sides of the triangle is constant. Rt. A DGC = rt. A DFC. (Why ?) .-. DF = DG. . . DE - DF^DE - DG =z EG, the ± distance between the two lis, BA and CH. Ex. 60. _ The sum of the perpendiculars dropped from any point in the base of an isosceles triangle to the legs is constant, and equal to the alti- tude upon one of the legs. Let PE and PB be the Js and BF the altitude. Draw PG X to BF. EPGF Is a parallelogram. (Why ?) .-. GF = PE. / \ff It remains to prove GB = PB. El The rt. A PGB = the rt. A BBP.. (Why ?) J- — p 'B Ex. 61. The sum of the perpendiculars dropped from any point within an equilateral triangle to the three sides is constant, and equal to the altitude. AD is the altitude, PE, PG, and PF the three per- pendiculars. Through P draw HK II to BC, meeting AD at M. Then MD = PE. (Why ?) PG + PF = AM (Ex. 60). Ex. 62. ABG and ABD are two triangles on the same base AB, and on the same side of it, the vertex of each triangle being without the other. H AC equals AD, show that BC cannot equal BD (§ 154). ^ Ex. 63. The sum of the lines which join a point within a triangle to the three vertices is less than the perimeter, but greater than half the perimeter. Ex. 64. If from any point in the base of an isosceles tri- angle parallels to the legs are drawn, a parallelogram is formed whose perimeter is constant, being equal to the sum E. of the legs of the triangle. B D 72 BOOK I. PLANE GEOMETRY. Ex. 65. The bisector of the vertical angle A ot a, tri- angle ABC, and the bisectors of the exterior angles at the base formed by producing the sides AB and AC, meet in a point which is equidistant from the base and the sides produced (§ 162). Ex. 66. If the bisectors of the base angles of a triangle are drawn, and through their point of intersection a line is drawn parallel to the base, the length of this parallel between the sides is equal to the sum of the segments of the sides between the parallel and the base. zsoB = zoBC = zob:e. ..be = bo. Ex. 67. The bisector of the vertical angle of a triangle makes with the perpendicular from the vertex to the base an angle equal ^ to half the difference of the base angles. Let Z B he greater than Z A. ZDCE = 90°-ZA ~ ZACD. ZACD = QO'' -iZA-iZB. -^ DE .■.ZDCE = 90°-ZA - (9Q°-iZA - iZB) = \ZB - \ZA. Ex. 68. If the diagonals of a quadrilateral bisect each -^ other, the figure is a parallelogram. Prove AAOB = ACOD. Ex. 69. The diagonals of a rectangle are equal. Prove A ABC = A BAD. -i Ex. 70. If the diagonals of a parallelogram are n equal, the figure is a rectangle. Ex. 71. The diagonals of a rhombus are perpendicular to each other, and bisect the angles of the rhombus. Ex. 72. The diagonals of a square are perpendicular to each other, and bisect the angles of the square. Ex. 73. Lines from two opposite vertices of a parallelogram to the middle points of the opposite sides trisect the diagonal. „ q EBFB is a O (why ?), and DF is II to BB. g/ AM = MN, and MN = CN (§ 188). EXERCISES. 73 F B L H G C ) G ( <> F t E L E B Ex. 74. The lines joining the middle points of the sides of any quadri- lateral, taken in. order, enclose a parallelogram. Prove HG and MF W to AC ; and FG and FH II to BD (§ 189). Then EG and FF are each equal to iAC. C A Ex. 75. The lines ioining the middle points of the sides of a rhombus, taken in order, enclose a rectangle. (Proof similar to that of Ex. 74.) Ex. 76. The lines joining the middle points of the sides of a rectangle (not a square) , taken in order, enclose a rhombus. Ex. 77. The Itues joining the middle points of the sides of a square, taken in order, enclose a square. Ex. 78. The lines joining the middle points of the sides of an isosceles trapezoid, taken in order, enclose a rhombus or a square. SHE and QFP drawn J. to AB are parallel. .-.^PQSB is a O, and by Const, is a rectangle or a square. .-. EFGS is a rhombus or a square (Exs. 76, 77). SB q O Q °^ A R E P B A B A Ex. 79. The bisectors of the angles of a rhomboid enclose a rectangle. Ex. 80. The bisectors of the angles of a rectangle enclose a square. Ex. 81. If two parallel lines are cut by a transversal, the bisectors of the interior angles form a rectangle. Ex. 82. The median of a trapezoid passes through the d q middle points of the two diagonals. The median ^i?" is II to ^B and bisects AB (§ 190). .-.it bisects DB. Likewise EF bisects BC and BD. 74 BOOK I. PLANE GEOMETRY. Ex. 83. The line joining the middle points of the diagonals of a trape- zoid is equal to half the difference of the bases. A BFG = A DFC. (Why ?) .. SF = iAG (% 189). CF = FG, DC = BG. .-. AG = AB- DC. .-. EF = i(AB - DC). n C 4E B I B Ex. 84. In an isosceles trapezoid each base makes equal angles with the legs. Draw CjB II to DB. CE = DB. (Why?) ZA = /.CEA,ZB = AGEA. A C and D have equal supplements. Ex. 85. If the angles at the base of a trapezoid are equal, the other angles are equal, and the trapezoid is isosceles. Ex. 86. In an isosceles trapezoid the opposite angles are supplementary. /.C = AD (Ex. 84). Ex. 87. The diagonals of an isosceles trapezoid are equal. Prove AACD = A BDC. Ex. 88. If the diagonals of a trapezoid are equal, the trapezoid is isosceles. C P Draw CE and DF _L to ^S. A ADF = A BCE. (Why ?) .: ADAF=ZCBA. A ABC = A BAD. A E F B F a Ex. 89. If from the diagonal DB, of a square ABCD, BE is cut off equal to BC, and EF is drawn perpendicu- lar to BD meeting DC at F, then DE is equal to EF and also to FC. /. EDF = 45°, and Z DFE = 45° ; and DE = EF. Kt. A BEF = rt. A BCF (§ 151) ; and EF = FC. Ex. 90. Two angles whose sides are perpendicular, each to each, are either equal or supplementary. Book II. THE CIRCLE. DEFINITIONS. 216. A circle is a portion of a plane bounded by a curved line, all points of which are equally distant from a point within called the centre. The bounding line is called the circumference of the circle. 217. A radius is a straight line from the centre to the cir- cumference; and a diameter is a straight line through the centre, with its ends in the circumference. By the definition of a circle, all its radii are equal. All its diameters are equal, since a diameter is equal to two radii. 218. Postulate. A circumference can be described from any point as a centre, with any given radius. 219. A secant is a straight line of unlimited length which intersects the circumference in two points; as, AD (Fig. 1). 220. A tangent is a straight line of unlimited length which has one point, and only one, in common with the circumference •,as,,BC (Fig- !)• In this case the circle is said to be tan- gent to the straight line. The common point is called the point of contact, or point of tangency. Fig. 1. 221. Two circles are tangent to each other, if both are tangent to a straight line at the same point ; and are said to be tangent internally or externally, according as one circle lies wholly within or without the other. 75 76 BOOK II. PLANE GEOMETRY. 222. An arc is any part of the circumference ; as, 5 C (Fig. 3). Half a circumference is called a semicircumference. Two arcs are called conjugate arcs, if their sum is a circumference. 223. A chord is a straight line that has its extremities in the circumference; as, the straight line BC.(Fig. 3). f 224. A chord subteads two conjugate arcs. If the arcs are unequal, the less is called the minor arc, and the greater the major arc. A minor arc is generally called simply an arc. 225. A segment of a circle is a portion of the circle bounded by an arc and its chord (Fig. 2). 226. A semicircle is a segment equal to half the circle (Fig. 2). 227. A sector of a circle is a portion of the circle bounded by two radii and the arc which they intercept. The angle included by the radii is called the anffle of the sector (Fig. 2). 228. A quadrant is a sector equal to a quarter of the circle (Fig. 2). 229. An angle is called a central angle, if its vertex is at the centre and its sides are radii of the circle ; as, Z.AOD (Fig. 2). 230. An angle is called an inscribed angle, if its vertex is i"n. the circumference and its sides are chords ; as, Z.ABC (Fig. 3). An angle is inscribed in a segment, if its vertex is in the arc of the segment and its sides pass through the extremities of the arc. AKCS, CHORDS, AND TANGENTS. 77 231. A polygon is inscribed in a circle, if its sides are chords ; and a circle is circumscrihed about a polygon, if all the vertices of the polygon are in the circumference (Fig. 3). 232. A circle is inscribed in a polygon, if the sides of the polygon are tangent to the circle ; and a polygon is circum- scribed about a circle, if its sides are tangents (Fig. 4). 233. Two circles are equal, if they have equal radii. For they will coincide, if their centres are made to coincide. Conversely : Tivo equal circles have equal radii. 234. Two circles are concentric, if they have the same centre. ARCS, CHORDS, AND TANGENTS. Proposition I. Theorem. 235. A straight line cannot meet the circumference of a circle in more than two points. • Let HK be any line meeting the circumference HKM in H and K. To prove that UK cannot meet the circumference in any other point. Proof. If possible, let HK meet the circumference in P. Then the radii OH, OP, and OK are equal. § 217 .'. P does not lie in the straight line HK. § 102 .•. ITS' meets the circumference in only two points. q.e.d. 78 BOOK II. PLANE GEOMETRY. Proposition II. Theorem. 236. In the same circle or in equal circles, equal central angles intercept equal arcs ; and of two unequal central angles the greater intercepts the greater arc. In the equal circles whose centres are and 0', let the angles AOB and A'O'B' be equal, and angle AOC be greater than angle A'O'B'. To prove that 1. arc AB = arc A'B' ; 2. arc AC > arc A'B'. Proof. 1. Place the O A'B'P' on the Q ABP so that the /.A'O'B' shall coincide with its equal, the Z. AOB. Then A' falls on A, and B' on B. § 233 .•. arc A'B' coincides with arc AB. § 216 2. Since the Z^OC is greater than the Z. A'O'B', it is greater than the Z AOB, the equal of the Z A'O'B'. Therefore, OC falls without the Z AOB. .•. arc ^C > arc AB. Ax. 8 .•. arc ^C > arc A'B', the equal of arc AB. q.e.d. 237. Conversely : In the same circle or in equal circles, equal arcs subtend equal central angles ; and of tivo un- equal arcs the greater subtends the greater central angle. AECS, CHORDS, AND TANGENTS. 79 To prove that 1. A AOB = AA'O'B' ; 2. Z.AOC is greater than A A'O'B'. Proof. 1. Place the O A'B'F' on the ABP so that O'A' shall fall on its equal OA, and the arc A'B' on its equal AB. Then O'B' will coincide with OB. § 47 .■.ZA'0'B< = Z.AOB. §60 2. Since arc AC> A'B', it is greater than arc AB, the equal of A'B', and 05 will fall within the ZAOC. .-.ZAOC is greater than Z ^ 05. Ax. 8 .-. Z ^00 is greater than Z ^'0'5'. Q.E.D. 238. CoK 1. iw the same circle or in equal circles, two sec- tors that have equal angles are equal; two sectors that have unequal angles are unequal, and the greater sector has the greater angle. 239. CoR. 2. In the same circle or in equal circles, equal sectors have equal angles; and of two unequal sectors the greater has the greater angle. 240. Law of Converse Theorems. It was stated in § 32 that the con- verse of a theorem is not necessarily true. If, however, a theorem is in fact a group of three theorems, and if one of the hypotheses of the group muM he true, and no two of the conclusions can be true at the same time, then the converse of the theorem is necessarily true. Proposition II. is a group of three theorems. It asserts that the arc AB is equal to the arc A'B', if the angle AOB is equal to the angle A'CXB' ; that the arc AB is greater than the arc A'B', if the angle A OB is greater than the angle A' O'B'; that the arc AB is less than the arc A'B', if the angle AOB is less than the angle A' O'B'. One of these hypotheses must he true ; for the angle A OB must be equal to, greater than, or less than, the angle A'O'B'. No two of the conclusions can be true at the same time, for the arc AB cannot be both equal to and greater than the arc A'B' ; nor can it be both equal to and less than the arc A'B' ; nor both greater than and less than the arc A'B'. In such a case, the converse theorem is necessarily true, and no proof like that given in the text is required to establish it. 80 BOOK II. PLANE GEOMETRY. Proposition III. Theorem. 241. In the same circle or in equal, circles, equal arcs are subtended hy equal chords; and of two unequal arcs the greater is subtended by the greater chord. In the equal circles whose centres are and 0', let the arcs AB and A'B' be equal, and the arc AF greater than arc A'B'. To prove that 1. chord J JB = chord A'B' ; 2. chord AF > chord A'B'. Proof. Draw the radii OA, OB, OF, O'A', O'B'. 1. , The AAOB and A' O'B' are equal. § 143 For OA = O'A', and OB = O'B', § 233 {radii of equal circles), and Z ^ 05 = Z A' O'B', § 237 {in equal © equal arcs subtend equal central A). . . chord AB = chord A'B'. § 128 2. In the A AOF and A' O'B', OA = O'A', and OF = O'B'. § 233 But the /-AOF \s greater than the Z A' O'B', § 237 {in equal ©, the greater of two unequal arcs subtends the greater Z). .-. chord AF > chord A'B'. § 154 Q.E.D. 242. Cor. In the same circle or in equal circles, the greater of tivo unequal major arcs is subtended by the less chord. ARCS, CHORDS, AND TANGENTS. 81 PROPOsiTioif IV. Theorem. 243. Conversely: In the same circle or in equal circles, equal chords subtend equal arcs; and of two unequal chords the greater subtends the greater arc. In the equal circles whose centres are and 0', let the chords AB and A'B' be equal, and the chord AF greater than A'B'. To prove that 1. arc AB = arc A'B' ; 2. arc AF > arc A'B'. Proof. Draw the radii OA, OB, OF, O'A', O'B'. 1. The A OAB and O'A'B' are equal. § 150 For OA = O'A', and OB = O'B', § 233 and chord AB = chord A'B'. HyP- .■.AAOB = Z.A'0'B'. §128 .-. arc ^5 = arc A'B', § 236 (yn equal ® equal central A intercept equal arcs). 2. In the A OAF and OA'B', OA = O'A' and 0F= O'B'. § 233 But chord AF > chord A'B'. Hyp. .-.the Z.AOF is greater than the Z A' O'B'. § 156 .-. arc ^i^ > arc A'B', § 236 {in equal ® the greater central Z intercepts the greater arc), q. e. d. 244. CoR. In the same circle or in equal circles, the greater of two unequal chords subtends the less major arc 82 BOOK II. PLANE GEOMETRY. Proposition V. Theorem. 245. A diameter perpendicular to a chord bisects the chord and the arcs subtended by it. Let ES be a diameter perpendicular to the chord AB at ffl. To prove that AM = BM, AS = BS, and AE = BE. Proof. Draw OA and OB from 0, the centre of the circle. The rt. A 0AM and OBM are equal. For OM = OM, and OA = OB. ,-. AM = BM, and Z AOS = Z BOS. Likewise ZAOE =ZB OE. .-. AS = BS, and AE = BE. §161 Iden. §217 §128 §85 § 236 Q.E.D. 246. CoE. 1. A diameter bisects the circumference and the circle. 247. Cor. 2. A diameter which bisects a chord is perpen- dicular to it. 248. Cor. 3. The perpendicular bisector of a chord passes through the centre of the circle, and bisects the arcs of the chord. ARCS, CHORDS, AND TANGENTS. 83 Pkoposition VI. Theorem. 249. In the same circle or in equal circles, equal chords are equally distant from the centre. Conveksely : Chords equally distant from the centre are equal. Let AB and CF be equal chords of the circle ABFC. To prove that AB and CF are equidistant from the centre 0. Proof. Draw OP ± to AB, Off±to CF, and join OA and OC. OF bisects AB, and Off bisects CF. § 245 The rt. A OF A and OffC are equal. § 151 For AP = Cff, Ax. 7 and OA = OC. § 217 Hence, OF = Off. § 128 .•. AB and CF are equidistant from O. Conversely : Let OP = OH. To prove AB = CF. Proof. The rt. A OPA and OffC are equal. § 151 For OA = OC, § 217 and OF = Off. Hyp. Hence, AF = Cff. § 128 .-. AB = CF. Ax. 6 Q.E.D. 84 BOOK II. PLANE GEOMETRY. Proposition VII. Theoeem. 250. In the same circle or in equal circles, if two chords are unequal, they are unequally distant from the centre; and the greater chord is at the less distance. In the circle whose centre is 0, let the chords AB and CD be unegual, and AB the greater; and let 0£ be perpendicular to AB and OF perpendicular to CD. To prove that OE < OF. Proof. Suppose AG drawn equal to CD, and OH l.i,o AG. Draw EH. OE bisects AB, and OH bisects AG. § 245 By hypothesis, AB > CD. .•.AB> AG, the equal of CD. .-. AE > AH. Ax. 7 .-. Z AHE is greater than Z AEH. § 152 .-. Z OHE, the complement of Z AHE, is less than Z OEH, the complement of Z AEH. Ax. 5 .-. OE < OH. § 163 But OH = OF. § 249 ■■■OE CD. Proof. Suppose AG drawn equal to CD, and OH'±to AG. Then OH=OF. §249 Hence, OE < OH. Draw EH. Z OHE is less than Z OEH. § 152 .". Z AHE, the complement of Z OHE, is greater than Z AEH, the complement of Z OEH. Ax. 5 .-. AE > AH. § J.53 But AE = i AB, and AH=iAG. § 245 .■.AB>AG. Ax. 6 But CD = AG. Const. .■.AB>CD. Q.E.D. 252. CoE. A diameter of a circle is greater than any ether chord. 86 BOOK II. PLANE GEOMETRY. Proposition IX. Theorem. 253. A straight line perpendicular to a radius at extremity is a tangent to the circle. Let MB be perpendicular to the radius OA at A. To prove that MB is a tangent to the circle. Proof. From draw any other line to MB, as OH. Then OH>OA. §97 .•. the point H is without the circle. § 216 Hence, every point, except A, of the line MB is without the circle, and therefore MB is a tangent to the circle at A. § 220 Q.E.D. 254. Cor. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. For OA is the shortest line from to MB, and is therefore ± to MB (§ 98) ; that is, MB is ± to OA. 255. Cor. 2. A perpendicular to a tangent at the point of contact passes through the centre of the circle. For a radius is _L to a tangent at the point of contact, and therefore a _L erected at the point of contact coincides with this radius and passes through the centre. 256. Cor. 3. A perpendicular from the centre of a circle to a tangent passes through the point of contact. ARCS, CHORDS, AND TANGENTS. 87 Peoposition X. Theorem. 257. Parallels intercept equal arcs on a circumfer- ence. F » „ A_ E B A FlO. 2. Case 1. Let AB (Fig. 1) be a tangent at F parallel to CD, a secant, To prove that arc CF = arc J)F. Proof. Suppose FF' drawn ± to AB. Then FF' is a diameter of the circle. And FF is also X to CD. .-. CF = DF, and CF' = BF'. Case 2. Let AB and CD (Fig. 2) be parallel secants. To prove that arc AC = arc BD. Proof. Suppose FF II tp CD and tangent to the circle at M. Then arc AM = arc 51f, Case 1 § 255 §107 §245 and arc CM = arc DM. .-. arc -4C = arc BD. Ax. 3 Case 3. Let AB and CD (Fig. 3) be parallel tangents at E and F. To prove that arc EGF — arc EHF. Proof. Suppose GH drawn II to AB. Then arc EQ = arc EH, Case 1 and are GF = arc HF. . wa EGF = dMQ, EHF. Ax. 2 Q.B.D. 88 BOOK II. PLANE GEOMETRY. Pbopositiost XI. Theorem. 258. Through three points not in a straight line one circumference, and ojily one, can he drawn. Let A, B, C be three points not in a straight line. To prove that one circumferenoe, and only one, can be drawn through A, B, and C. Proof. Draw AB and BC. At the middle points of AB and BC suppose Js erected. These Js wUl intersect at some point 0, since AB and BC are not in the same straight line. The point is in the perpendicular bisector of AB, and is therefore equidistant from A and B ; the point is also in the perpendicular bisector of BC, and is therefore equidistant from B and C. § 160 Therefore, is equidistant from A, B, and C ; and a cir- cumference described from as a centre, with a radius OA, will pass through the three given points. The centre of a circumference passing through the three points must be in both perpendiculars, and hence at their intersection. As two straight lines can intersect in only one point, is the centre of the only circumference that can pass through the three given points. q.rd. 259. CoR. Two circumferences can intersect in only two points. For, if two circumferences have three points common, they coincide and form one circumference. ARCS, CHORDS, AND TANGENTS. 89 260. Def. a tangent from an external point to a circle is the pai-t of the tangent between the external point and the point of contact. Peoposition XII. Theoeem. 261. The tangents to a circle drawn from an external point are equal, and make equal angles with the line joining the point to the centre. let AB and AC be tangents from A to the circle whose centre is 0, and let AO be the line joining A to the centre 0. To prove that AB = AC, and Z BAO = Z CAO. Proof. Draw OB and OC. AB is _L to OB, and AC ± to OC, § 254 (a tangent to a circle is _L to the radius drawn to the point of contact). The rt. A OAB and OAC are equal. § 151 For OA is common, and the radii OB and OC are equal. § 217 .■.AB = AC,&ndZBAO = ZCAO. §128 Q.E.D. 262. Dee. The line joining the centres of two circles is called the line of centres. 263. Dee. A tangent to two circles is called a common external tangent if it does not cut the line of centres, and a common internal tangent if it cuts the line of centres. 90 BOOK n. PLANE GEOMETRY. Proposition XIII. Theoeem. 264. If two circles intersect each other, the line of centres is perpendicular to their common chord at its middle point. ■-'A ^-^ y { \ ■■. H V Let C and C be the centres of the two circles, AB the common chord, and CC the line of centres. To prove that GC is _L to AB at its middle point. Proof. Draw CA, CB, C'A, and CB. CA = CB, and C'A = CB. § 217 .-. C and C" are two points, each equidistant from A and B. .". CC is the perpendicular bisector of AB. § 161 Q.E.D. Ex. 92. Describe the relative position of two circles if the line of centres : (1) is greater than the sum of the radii ; (2) is equal to the sum of the radii ; (3) is less than the sum hut greater than the difference of the radii ; (4) is equal to the diSerence of the radii ; (6) is less than the difierence of the radii. Illustrate each case by a figure. Ex. 93. The straight line drawn from the middle point of a chord to the middle point of its subtended arc is perpendicular to the chord. Ex. 94. The line which passes through the middle points of two par- allel chords passes through the centre of the circle. ARCS, CHORDS, AND TANGENTS. 91 Proposition XIV. Theorem. 265. If two circles are tangent to each other, the line of centres passes through the point of contact. Let the two circles, whose centres are C and C, be tangent to the straight line AB at 0, and CC the line of centres. To prove that is in the straight line CC. Proof. A ± to AB, drawn through the point 0, passes through the centres C and C", § 255 (a I. to a tangent at the point of contact passes through the centre of the circle). .'. the line CC, having two points in common with this ± must coincide with it. § 47 .'. is in the straight line CC q.e.d. Ex. 95. Describe the relative position of two circles if they may have: (1) two common external and two common internal tangents ; (2) two common external tangents and one common internal tangent ; (3) two common external tangents and no common internal tangent ; (4) one common external and no common internal tangent ; (5) no common tangent. Illustrate each case by a figure. Ex. 96. The line drawn from the centre of a circle to the point of inter- section of two tangents is the perpendicular bisector of the chord joining the points of contact. 92 BOOK II. PLANE GEOMETRY. MEASUREMENT. 266. To measure a quantity of any kind is to find the number of times it contains a known quantity of the same kind, called the unit of measure. The number which shows the number of times a quantity contains the unit of measure is called the numerical measure of that quantity. 267. No quantity is great or small except by comparison with another quantity of the same kind. This comparison is made by finding the numerical measures of the two quantities in terms of a common unit, and then dividing one of the measures by the other. The quotient is called their ratio. In other words the ratio of two quantities of the same kind is the ratio of their numeri- cal measures expressed in terms of a common unit. a . The ratio of a to i is written a:b, or y • 268. Two quantities that can be expressed in integers in terms of a common unit are said to be commensurable, and thb exact value of their ratio can be found. The common unit is called their common measure, and each quantity is called a multiple of this common measure. Thus, a common measure of 2\ feet and 3| feet is | of a foot, which is contained 15 times in 2| feet, and 22 times in 3| feet. Hence, 2| feet and 3| feet are multiples of | of a foot, since 2 J feet may be obtained by taking |^ of a foot 15 times, and 3| feet by taking I of a foot 22 times. Tlie ratio of 2^- feet to 3f feet is expressed by the fraction \ %. 269. Two quantities of the same kind that cannot both be expressed in integers in terms of a common unit, are said to be incommensurable, and the exact value of their ratio cannot be found. But by taking the unit sufficiently small, an approxi- mate value can be found that shall differ from the true value of the ratio by less than any assigned value, however small. THEORY OP LIMITS. 93 Thus, suppose the ratio, -=V2. Now V2 = 1.41421366 • • •, a value greater than 1.414213, but less than 1.414214. If, then, a millionth part of b is taken as the unit of measure, the value of -r lies between 1.414213 and 1.414214, and therefore b diifers from either of these values by less than 0.000001. By carrying the decimal further, an approximate value may be found that will differ from the true value of the ratio by less than a billionth, a trillionth, or any other assigned value. In general, if -r > — hut < ; then the error in taking b n n either of these values for — is less than —> the difference b n between these two fractions. But by increasing n indefi- nitely, - can be decreased indefinitely, and a value of the ratio can be found within any required degree of accuracy. 270. The ratio of two incommensurable quantities is called an incommensurable ratio ; and is a fixed value which its suc- cessive approximate values constantly approach. THE THEORY OF LIMITS. 271. When a quantity is regarded as having a fixed value throughout the same discussion, it is called a constant ; but when it is regarded, under the conditions imposed upon it, as having different successive values, it is called a variable. If a variable, by having different successive values, can be made to differ from a given constant by less than any assigned value, however small, but cannot be made absolutely equal to the constant, that constant is called the limit of the variable, and the variable is said to approach the constant as its limit. 94 BOOK IL PLANE GEOMETRY, 273. Suppose a point to move from A toward B, undei the conditions that the ^ m x" m" b first second it shall I 1 • move one half the distance from A to B, that is, to M; the next second, one half the remaining distance, that is, to M'; and so on indefinitely. Then it is evident that the moving point may approach as near to B as we choose, hut will never arrive at B. For, how- ever near it may be to B at any instant, the next second it will pass over half the distance still remaining; it must, therefore, approach nearer to B, since half the distance still remaining is some distance, but will not reach B, since half the distance still remaining is not the whole distance. Hence, the distance from A to the moving point is an increasing variable, which indefinitely approaches the constant AB as its limit ; and the distance from the moving point to 5 is a decreasing variable, which indefinitely approaches the constant zero as its limit. 273. Again, suppose a square ABCD inscribed in a circle, and E, F, H, K the middle points of the arcs subtended by the sides of the square. If we draw the lines AE, EB, BF, etc., we shall have an inscri bed polygon of double the number of sides of the square. The length of the perimeter of this K{ polygon, represented by the dotted lines, is greater than that of the square, since two sides replace each side of the square and form with it a triangle, and two sides of a triangle are together greater than the third side; but less than the length of the circumference, for it is made up of straight lines, each one of which is less than the pai't of the circumference between its extremities. THEORY OF LIMITS. 95 By continually doubling the number of sides of each result- ing inscribed figure, the length of the perimeter -will increase with the increase of the number of sides, but wlU. not become equal to the length of the circumference. The difference between the perimeter of the inscribed poly- gon and the circumference of the circle can be made less than any assigned value, but cannot be made equal to zero. The length of the circumference is, therefore, the limit of the length of the perimeter as the number of sides of the inscribed figure is indefinitely increased. § 271 274. Consider the decimal 0.333 • • • which may be written A + TT7 + TtftrTy + • • ■ The Talue of each fraction after the first is one tenth of the preceding fraction, and by continuing the series we shall reach a fraction less than any assigned value, that is, the values of the successive fractions approach zero as a limit. The sum, of these fractions is less than -J; but the more terms we take, the nearer does the sum approach ^ as a limit. 275. Test for a limit. In order to prove that a variable approaches a constant as a limit, it is necessary to prove that the difference between the variable and the constant : 1. Can he made less than any assigned value, however small. 2. Cannot be made absolutely equal to zero. 276. Theorem. If the limit of a variable x is zero, then the limit of kx, the product of the variable by any finite constant k, is zero. 1. Let q be any assigned quantity, however small. Then | is not 0. Hence x, which may differ as little as we please from 0, may he taken less than | > and then kx will be less than q. 2. Since x cannot be 0, lex cannot be 0. Therefore, the limit of kx = 0. § 275 96 BOOK IL PLANE GEOMETRY. 277. Cor. If the limit of a variable x is zero, then the limit of the quotient of the variable by any finite constant k, is also zero. X 1 For - = - X K, which by § 276 can be made less than any assigned value, however small, but cannot be made equal to zero. 278. Theorem. The limit of the sum of a finite number of variables x, y, z, • • ■ is equal to the sum of their respective limits a, b, c, ■ ■ ■. Let d, d', d", • • ■ denote the differences between x, y, z,-- ■ and a, b, c, • • • , respectively. Then d + d' + d" + ■ ■■ can be made less than any assigned quantity q. For, if d, d', d", ■ • ■ are n in number and d is the largest, d + d' + d" + ■■■ A POM = A POF (§ 14.3). .-. OE = OF. A POE' Ex. 131. If ABC is an inscribed equilateral triangle and P is any point in the arc BC, then PA = PB + PC. Take PM = PB. A ABM = A CBP (§ 143) and AM = PC. Ex. 132. The tangents drawn through the vertices of an inscribed rectangle, which is not a square, enclose a rhombus. a si Ex. 133. The bisectors of the angles included by the opposite sides (produced) of an mscribed quadrilateral intersect at right angles. Arc AF - arc BM = arc DF - arc CM and arc Afl — arc DN = arc BH — arc CN. .-. arc FH + arc MN = arc HM + arc FN. . . Z FIH = Z RIM. Discussion. This problem is impossible, if any two sides of the quadri- lateral are parallel. 112 BOOK n. PLANE GEOMETEY. PROBLEMS OP CONSTRUCTION. Note. Hitherto we have supposed the figures constructed. We now proceed to explain the methods of constructing sunple problems, and after- wai-ds to apply these methods to the solution of more difficult problems. Proposition XXI. Pkoblbm. 300. To let fall a perpendicular upon a given line from a given external point. A \ H "-- M ■-'K " . Let AB be the given straight line, and C the given external point. To let fall a A- to the line AB from the point C. Prom C as a centre, with a radius sufB.ciently great, describe an arc cutting AB in two points, H and K. Prom H and K as centres, with equal radii greater than ^ HK, describe two arcs intersecting at 0. Draw CO, and produce it to meet AB at M. CM is the -L required. Proof. Since C and are two points each equidistant from H and K, they determine a ± to HK at its middle point. § 161 Q.E.F. Note. Given lines of the figures are represented by full lines, resulting lines by long-dashed, and auxiliary lines by short-dashed lines. PROBLEMS OF CONSTRUCTION. 113 Proposition XXII. Problem. 301. At a given point in a straight line, to erect a perpendicular to that line. :^ \/ J/ Fig. 1. -Bid. a. 1. Let be the given point in AC. Fig. 1. Take OH and OB equal. From H and B as centres, with equal radii greater than OB, describe two arcs intersecting at B. Join OR. Then the line OR is the _L required. Proof. and R, two points each equidistant from H and B, determine the perpendicular bisector of HB. § 161 2. Let B be the given point. Fig. 2. Take any point C without AB; and from C as a centre, with the distance CB as a radius, describe an arc intersecting AB at E. Draw EC, and prolong it to meet the arc again at D. Join BD, and BD is the _L required. Proof. The A Bis a, right angle. § 290 .■.BD\s±to AB. Q.E.F. Discussion. The point C must be so taken that it will not be in the required perpendicular. 114 BOOK II. PLANE GEOMETRY. Proposition XXIII. Pkoblem. 302. To bisect a given straight line. f D To bisect the given straight line AB. From A and B as centres, with equal radii greater than ^ AB, describe arcs intersecting at C and E. Join CE. Then CE bisects AB. Proposition XXIV. Problem. 303. To bisect a given arc. § 161 Q. E. F. To bisect the given arc AB. Draw the chord AB. From A and B as centres, with equal radii greater than ^ AB, describe arcs intersecting at D and E. PROBLEMS OP CONSTRUCTIOK. 115 Draw DE. Then DE is the ± bisector of the chord AB. § 161 .-. DE bisects the arc ACB. § 248 Q. E. F. Peoposition XXV. Problem. 304. To bisect a given angle. lex. AEB be the given angle. Prom E as a centre, with any radius, as EA, describe an arc cutting the sides of the Z ^ at ^ and B. Prom A and B as centres, with equal radii greater than half the distance from A to B, describe two arcs intersecting at D. Draw DE. Then DE bisects the arc AB at C. DE bisects the angle E. § 303 § 237 Q. E. F. Ex. 134. To construct an angle of 4-5° ; of 1.35°. Ex. 135. To construct an equilateral triangle, having given one side. Ex. 136. To construct an angle of 60°; of 150°. Ex. 137. To trisect a right angle. 116 BOOK II. PLANE GEOMETRY. Proposition XXVI. Peoblem. 305. At a given point in a given straight line, to construct an angle equal to a given angle. At C in the line CM, construct an angle equal to the given angle A. From ^ as a centre, with any radius, AIS, describe an are cutting the sides of the Z ^ at ^ and F. Prom C as a centre, with a radius equal to AE, describe an arc HG cutting CM at H. From ^ as a centre, with a radius equal to the chord EF, describe an arc intersecting the arc HG at 0. Draw CO, and Z. HCO is the required angle. Why ? Q.E.F. Proposition XXVII. Problem. 306. To draiv a straight line parallel to a given straight line through a given external point. E / rr^ Cj :^- '~>-: i\ D i Let AB be the given line, and C the given point. PROBLEMS OF CONSTRUCTION. 117 Draw BCD, making any convenient Z EDB. At the point C construct Z EOF equal to Z EDB. § 305 Then the line HCF is II to AB. Why ? Q.E.F. Proposition XXVIII. Problem. 307. To divide a given straight line into a given number of equal parts. / / c--- ""-0 Let AB be the given straight line. From A draw the line AO, making any convenient angle with AB. Take any convenient length, and apply it to ^0 as many times as the line AB is to be divided into parts. From C, the last point thus found on AO, draw CB. Through the points of division on ^0 draw parallels to the line CB. § 306 These lines will divide AB into equal parts. § 187 Q.E.F. Ex. 138. To construct an equilateral triangle, having given the per- imeter. Ex. 139. To divide a line into four equal parts hy two different methods. Ex. 140. Through a given point to draw a line which shall make equal anglis with the two sides of a given angle. Through the given point draw a X to the bisector of the given Z. Ex. 141. To draw a line through a given point, so that it shall form with, the sides of -a given, angle au isosceles triangle (Ex. 140). 118 BOOK n. PLANE GEOMETRY. Proposition XXIX. Problem. 3Q8. To find the third angle of a triangle when iioo of the angles are given. B I I t E- Let A and B be the two given angles. At any point H in any line EF, ^ construct Z a equal to Z A, and Z h equal to /. B. § 305 Then Z c is the Z required. Why ? Q.E.F. Proposition XXX. Problem. 309. To construct a triangle when two sides and the included angle are given. X" A ; c B Let b and c be the two sides of the triangle and E the included angle. Take AB equal to the side c. At A, construct Z BAD equal to the given AE. § 305 PROBLEMS OF CONSTRUCTION. 119 On AB take AC equal to b, and draw CB. Then A ACB is the A required. q.e.f. Proposition XXXI. Problem. 310. To construct a triangle when a side and two angles of the triangle are given. Let c be the given side, A and B the given angles. Take JS-C equal to the side c. At E construct the Z CEH equal to Z ^. § 305 At C construct the Z ECK equal to Z B. Produce EH and CK untU they intersect at 0. Then A COE is the A required. q.e.f. Rkmakk. If one of the given angles is opposite to the given side, find the third angle by § 308, and proceed as above. Discussion. The problem is impossible when the two given angles are together equal to or greater than two right angles. Ex. 142. To construct an equilateral triangle, having given the altitude. To construct an isosceles triangle, having given : Ex. 143. The base and the altitude. Ex. 144. The altitude and one of the legs. Ex. 145. The angle at the vertex and the altitude. 120 BOOK n. PLANE GEOMETRY. Pkoposition XXXII. Problem. 311. To construct a triangle when two sides and the angle opposite one of them are given. by a \a \ -All Let a and b be the given sides, and A the angle opposite a. Case 1. If a is less than h. Construct Z. DAE equal to the given /.A. § 305 On AD take AB equal to b. Prom i? as a centre, witli a radius equal to a, describe an arc intersecting the line AE at G and C". Draw BG and BG'. Then both the A ABG and ABG fulfil the conditions, and hence we have two constructions. 2) This is called the ambiguous case. B,--'' i" H Discussion. If the side a is equal to y \a the _L BH, the arc described from B ^ A. will touch AE, and there will be but one construction, the right A ABH. If the given side a is less than the By'' ± from B, the arc described from B ^y^ \a will not intersect or touch AE, and /' --• hence the problem is impossible. — ;^ tHOBLEMS Of CONSTRUCTION. 121 If the Z.Aia right or obtuse, the problem is impossible ; for the side opposite a right or obtuse angle is the greatest side. § 163 Case 2. If a is equal to b. If the Z. A is acute, and a = b, the arc described from B as a centre, and with a radius equal to a, will ^ cut the -line AH! at the points A and C. B/ There is therefore but one solution: the . y \ •s / isosceles A ABC. -A"--^ ---'0 E Discussion. If the Z. A is right or obtuse, the problem is impossible ; for equal sides of a A have equal A opposite them, and a A cannot have two right A or two obtuse A. Case 3. If a is greater than b. If the given A A is acute, the arc described from B will cut the line HD on opposite sides of A, at C and C". The A ABC answers the required conditions, but the A ABC does not, for it does not contain R the acute A A. There is then only one ^ j/''-^ "'--,/ solution ; namely, the A ABC. C^\4 yb i6^« If the A A is right, the arc described B from B cuts the line HD on opposite a/ sides of A, and we have two equal right ^ \ / i \/ _ A which fulfil the required conditions. ^ ' ^ If the A A is obtuse, the arc described \b from B cuts the line ED on opposite a^ \\a sides of A, at the points C and C". The ^ \/ \ y p A ABC answers the requii-ed conditions, but the A ABC does not, for it does not contain the obtuse A A. There is then only one solution; namely, the A ABC. Q.E.F. 122 BOOK II. PLANE GEOMETRY. Proposition XXXIII. Problem. 312. To construct a triangle tvhen the three sides of the triangle are given. by Let the three sides be c, a, and b. Take AB equal to c. From ^ as a centre, with a radius equal to b, describe an arc. Prom B as & centre, with a radius equal to a, describe an arc, intersecting the other arc at C. Draw CA and CB. A CAB is the A required. q e p Discussion. The problem is impossible when one side is equal to or greater than the sum of the other two sides. Proposition XXXIV. Problem. 313. To construct a parallelogram when two sides and the included angle are given. /■ / A Let m and o be the two sides, and C the included angle. Take AB equal to o. At A construct Z BAB equal to Z C. § 305 PROBLEMS OF CONSTRUCTION. 123 Take AH equal to m. From ^as a centre, mth a radius equal to o, describe an arc, and from 5 as a centre, with a radius equal to m, describe an arc, intersecting the other arc at E; and draw EH and EB. The' quadrilateral ABEH is the O required. § 182 0. E.F. Proposition XXXV. Pboblbm. 314. To circumscribe a circle about a given triangle. Let ABC be the given triangle. Bisect AB and BO. § 302 At E and D, the points of bisection, erect Js. § 301 Since BC is not the prolongation of AB, these Js will inter- sect at some point O. From 0, with a radius equal to OB, describe a circle. The O ABC is the O required. Proof. The point is equidistant from A and B, and also is equidistant from B and C. § 160 .•. the point is equidistant from A, B, and 0, and a O described from as a centre, with a radius equal to OB, wiU pass through the vertices A, B, and C. q.e.f. The same construction serves to describe a circumference which shall pass through three points not in the same straight line ; also to find the centre of a given circle or of a given arc. Note. Tlie point is called the circum-centre of the triangle. 124 BOOK II. PLANE GEOMETRY. Pkoposition XXXVI. Problem. 315. To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect tte A A and C. § 304 From E, the intersection of the bisectors, draw EHl. to the side AC. § 300 From JS as centre, with radius EH, describe the O KUM. The O KHM is the required. Proof. Since E is in the bisector of the Z. A, it is equidistant from the sides AB and A C ; and since E is in the bisector of the Z. C, it is equidistant from the sides AC and BC. § 162 .".a described from E as centre, with a radius equal to EH, will touch the sides of the A and be inscribed in it. q.e.f. Note. The point E is called the in-centre of the triangle. 316. The intersections of the bisectors of the exterior angles of a triangle are the centres of three circles, each of which will touch one side of the triangle, and the two other sides produced. These three circles are called escribed circles ; and their centres are called the ex-centres of the triangle. PROBLEMS OF CONSTRUCTION. 125 Pkoposition XXXVII. Problem. 317. Through a given point, to draio a tangent to a given circle. — M ;J^.--" -^"E Case 1. When the given point is on the circumference. Let C be the given point on the circumference whose centre is 0. From the centre draw the radius OG. Through C draw AMI. to OC. § 301 Then AM is the tangent required. § 263 Case 2. When the given point is without the circle. Let be the centre of the given circle, E the given point. Draw OE. On OE as a diameter, describe a circumference intersecting the given circumference at the points M and H. Draw OJf and ^If. Then EM is the tangent required. Proof. Z OME is a right angle. § 290 .-. EM is tangent to the circle at M. § 263 In like manner, we may prove EH tangent to the given O. Q.E.F. Ex. 146. To draw a tangent to a given circle, so that it shall be parallel to a given straight line. 126 BOOK II. PLANE GEOMETRY. Proposition XXXVIII. Problem. 318. Upon a given straight line, to describe a seg- ment of a circle in loliich a given angle may he inscribed. /E Let AB be the given line, and M the given angle. Construct the Z ABE equal to the Z. M. § 305 Bisect the line AB by the _L OF. § 302 Prom the point B draw BO ± to JEB. ' § 301 Prom 0, the point of intersection of FO and BO, &s a cen- tre, with a radius equal to OB, describe a circumference. The segment AKB is the segment required. Proof. The point is equidistant from A and B. § 160 .•. the circumference will pass through A. But BF is J. to OB. Const. ■.-. BF is tangent to the O, § 253 (a straight line ±to a radius at its extremity is tangent to the G). . . Z ABF is measured by ^ arc AB, § 295 {being an Z formed by a tangent and a chord). But any Z as Z /f inscribed in the segment AKB is meas- ured by i arc AB. § 289 .•. the Z M may be inscribed in the segment AKB. q.e.f. SOLUTION OF PROBLEMS. 127 SOLUTION OF PROBLEMS. 319. If a problem is so simple that the solution is obvious from a known theorem, we have only to make the construction according to the theorem, and then give a synthetic proof, if a proof is necessary, that the construction is correct, as in the examples of the fundamental problems already given. 320. But problems are usually of a more difficult type. The application of known theorems to their solution is not immediate, and often far from obvious. To discover the mode of application is the fii'st and most difficult part of the solu- tion. The best way to attack such problems is by a method resembling the analytic proof of a theorem, called the analysis of the problem. 1. Suppose the construction made, and let the figure repre- sent all parts concerned, both given and required. 2. Study the relations among the parts with the aid of known theorems, and try to find some relation that will sug- gest the construction. 3. If this attempt fails, introduce new relations by drawing auxiliary lines, and study the new relations. If this attempt fails, make a new trial, and so on till a clue to the right con- struction is found. 321. A problem is determinate if it has a definite number of solutions, indeterminate if it has an indefinite number of solutions, and impossible if it has no solution. A problem is sometimes determinate for certain relative positions or magni- tudes of the given parts, and indeterminate for other positions or magnitudes of the given parts. 322. The discussion of a problem consists in examining the problem with reference to all possible conditions, and in deter- mining the conditions necessary for its solution. 128 BOOK n. PLANE GEOMETftV. Ex.147. Problem. To construct a circle that shall pass through a given point and cut chords of a given length from two parallels. Analysis. Suppose the problem solved. Let A be the given point, BC and DE the given parallels, MN the given length, and the centre of the required circle. Since the circle cuts equal chords from two parallels its centre must be equidistant from them. Therefore, one locus for is FG II to BC and equidistant from BG and DE. Dravi' the J. bisector of MN, cutting FG in P. PM is the radius of the circle required. With A as centre and radius PM describe an arc cutting FG at 0. Then is the centre of the required circle. Discussion. The problem is impossible if the distance from A to FG is gi-eater than PM. D E F F \ „^A / ~&\ B M N c Ex.148. Problem. Toconstruct a triangle, having given th^ perimeter, one angle, and the altitude from the vertex of the given angle. Analysis. Suppose the problem solved, and let ABC be the A required, A CB the given Z, and CD the given altitude. Produce AB both ways, and take AE = AC, and BF=BC, then EF = the given perim- eter. Join CE and CF, form- ing the isosceles k. CAE and CBF. IntheAE^CJF, /.E Jr AF + Z ECF = 180° (why ?), but Z ECF = Z EGA D/B ZFCB + ZACB. Since ZE = Z EGA and / F = Z FCB, we have ZECF = ZE + ZF + ZACB. ..'iZE + 2ZF + ZACB = \?,Q°. .. ZE + ZF + iZACB = QO°, and ZE + ZF =Q0° - iZACB. By substitution, Z ECF = m° + i Z ACB. .-. Z ECF is known. Construction. To find the point C, construct on EF a segment that will contain the ZECF (§ 318), and draw a parallel to EF at the dis- tance CD, the given altitude. To find the points A and B, draw the ± bisectors of the lines GE and CF, and the points A and B will be vertices of the required A. Why ? EXERCISES. 129 PEOBLEMS OF CONSTETJCTION. Ex. 149. Find the locus of a point at a given distance from a given ciromnference. Find tlie locus of ttie centre of a circle : Ex. 150. "Which has a given radius r and passes through a given point P. Ex. 151. "Which has a given radius r and touches a given line AB. Ex. 152. Which passes through two given poiuts P and Q. Ex. 153. "Which touches a given straight line AB at a given point P. Ex. 154. "Which touches each of two given parallels. Ex. 155. "Which touches each of two given intersecting lines. Ex. 156. To find in a giveu line a point X which is equidistant from two given points. The required point is the intersection of the given line with the perpen- dicular bisector of the line joining the two given points (§ 160). Ex. 157. To find a point X equidistant from three given points. Ex. 158. To find a point X equidistant from two given ,--- points and at a giveu distance from a third given point. 'y^^ •\--7—R Ex. 159. To construct a circle which has a given radius }? and passes through two given points. •'' '^ Ex. 160. To find a point X at given distances from two given points. Ex. 161. To construct a circle which has its centre in a given line and passes through two given points. Ex. 162. To find a point X equidistant from two given points and also equidistant from two given intersecting lines (§§ 160 and 162). Ex. 163. To find a point X equidistant from two given points and also equidistant from two given parallel lines. Ex. 164. To find a point X equidistant from two given intersecting lines and also equidistant from two given parallels. Ex. 165. To find a point X equidistant from two given g -ij— .. „ intersecting lines and at a given distance from a given point. -^ / j p Ex. 166. To find a point X which lies in one side of a given triangle and is equidistant from the other two sides. 130 BOOK II. PLANE GEOMETRY. x; 1 ir n ' ' n s Ex. 167. A straight railway passes two miles from a town. A place is four miles from the towa and one mile from the railway. To find by construction the places that answer this description. Ex. 168. In a triangle ABC, to draw DE parallel to the base BO, cutting the sides of the triangle in D and E, so that DM shall equal DB + EC (§ 162). ^ p Ex. 169. To di-aw through two sides of a triangle a line parallel to the third side so that the part intercepted between p/\^;;#~ the sides shall have a given length. Take BD = d. Ex. 170. Prove that the locus of the vertex of a right triangle, having a given hypotenuse as base, is the cu-cumference described upon the given hypotenuse as diameter (§ 290). Ex. 171. Prove that the locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms with the base a segment capable of containing the given angle (§ 318). Ex. 172. Find the locus of the middle point of a chord of a given length that can be drawn in a given circle. Ex. 173. Find the locus of the middle point of a chord drawn from a given point in a given circumference. Ex. 174. Find the locus of the middle point of a straight line drawn from a given exterior point to a given circumference. E A B '^ A' '' Ex. 175. A straight line moves so that it remains parallel to a given line, and touches at one end a given circumference. Find the locus of the other end. Ex. 176. A straight rod moves so that its ends con- , stantly touch two fixed rods which are perpendicular to -{- each other. Find the locus of its middle point. ^^ EXERCISES. 131 Ex. 177. In a given circle let AOB be a diameter, OC any radius, CD the i^erpeudicular from C to AB. Upon OC take OM equal to CD. Find tlie locus of tlie point M as OC turns about 0. _ p Ex. 178. To construct an equilateral triangle, having /' ,''',-■ given the radius of the circumscribed circle. ^ f--,^ o'\ To construct an isosceles triangle, having given : "~ — '2 Ex. 179. The angle at the vertex and the base (§ 160 and § 318). Ex. 180. The base and the radius of the circumscribed circle. Ex. 181. The base and the radius of the inscribed cncle. Ex. 182. The perimeter and the altitude. c Let ABC be the A required, MF the given perim- ,^'/l\\x eter. The altitude CD passes through the middle ^^ \ / i \ / ^^-^ of EF, and the ^ ABC, BFC are isosceles. £' Ar~D^ F To construct a right triangle, having given : Ex. 183. The hypotenuse and one leg. Ex. 184. One leg and the altitude upon the hypotenuse. Ex. 185. The median and the altitude drawn from the vertex of the right angle. Ex. 186. The hypotenuse and the altitude upon the hypotenuse. Ex. 187. The radius of the inscribed circle and one leg. Ex. 188. The radius of the inscribed circle and an acute angle. Ex. 189. An acute angle and the sum of the legs. 4-, Ex. 190. An acute angle and the difference of the legs. / „ '/^vX p \ Ex. 191. To construct an equilateral triangle, having I / A''^''j\'' given the radius of the inscribed circle. V^"----- ■yc To construct a triangle, having given : Ex. 192. The base, the altitude, and an angle at the base. Ex. 193. The base, the altitude, and the Z at the vertex. Ex. 194. The base, the corresponding median, and the Z at the vertex, Ex. 195. The perimeter and the angles. Ex. 196. One side, an adjacent Z, and the sum of the other sides. 182 BOOK II. PLANE GEOMETRY. To construct a triangle, having given : Ex. 197. One side, an adjacent Z, and the difference of the other sides. Ex. 198. The sum of tvfo sides and the angles. Ex. 199. One side, an adjacent Z, and the radius of the circumscribed circle. Ex. 200. The angles and the radius of the circumscribed circle. Ex. 201. The angles and the radius of the inscribed circle. Ex. 202. An angle, and the bisector and the altitude dravm from the vertex of the given angle. Ex. 203. Two sides and the median corresponding to the other side. Ex. 204. The three medians. To construct a square, having given : Ex. 205. The diagonal. Ex. 206. The sum of the diagonal and one side. d. Let ABCD be the square required, CA the diagonal. ^ ^/' \ Produce GA, making A^ = AB. /^ ABG and ABS are -"v.- V /O isosceles vm^ ABAC = ABCA = 45°. "% Ex. 207. Given two perpendiculars, AB and CI), intersecting in 0, and a straight line intersectiug these perpendiculars in E and F; to construct a square, one of whose angles shall coincide with one of the right angles at 0, and the vertex of the oppo- site angle of the square shall lie in HF. (Two solu- tions. ) To construct a rectangle, having given : Ex. 208. One side and the angle between the diagonals. Ex. 209. The perimeter and the diagonal. Ex. 210. The perimeter and the angle between the diagonals. Ex. 211. The difference of two adjacent sides and the angle between the diagonals. To construct a rhombus, having given : Ex. 212. The two diagonals. Ex. 213. One side and the radius of the inscribed circle. F \ N V,--' ' \m\ I ' ~ A i a! « ^H EXEKCISES. 133 Ex. 214. One angle and the radius of the mscribed circle. Ex. 215. One angle and one of the diagonals. To construct a rhomboid, having given : Ex. 216. One side and the two diagonals. Ex. 217. The diagonals and the angle between them. Ex. 218. One side, one angle, and one diagonal. Ex. 219. The base, the altitude, and one angle. To construct an isosceles trapezoid, having given : Ex. 220. The bases and one angle. Ex. 221. The bases and the altitude. Ex. 222. The bases and the diagonal. Ex. 223. The bases and the radius of the ch-cumscribed circle. Let ABCD be the isosceles trapezoid required, the z>^: centre of the circumscribed G. A diameter X to ^B is i. to // '/ 7l\ i-L ^\ \E q.'B CD, and bisects both AB and CD. Draw CG II to FE. Then EG = FG = iDC. " ----1--'- To construct a trapezoid, having given : Ex. 224. The four sides. Ex. 225. The two bases and the two diagonals. Ex. 226. The bases, one diagonal, and ^he Z between the diagonals. To construct a circle which has the radius r and which also : Ex. 227. Touches each of two intersecting lines AB and CD. Ex. 228. Touches a given line AB and a given circle K. Ex. 229. Passes through a given point P and touches a given line AB. Ex. 230. Passes through a given point P and touches a given circle K. To construct a circle which shall : Ex. 231. Touch two given parallels and pass through a. given point P. Ex. 232. Touch three given lines two of which are parallel. Ex. 233. Touch a given line AB&tP and pass through a given point Q. Ex. 234. Touch a given circle at P and pass through a given point Q. Ez. 235. Touch two given lines and touch one of them at a given point P. 134 BOOK II. PLANE GEOMETRY. Ex. 236. Touch a given line and toucli a given circle at a point P. Ex. 237. Touch a given line AB at P and also touch a given circle. Ex. 238. To inscribe a circle in a given sector. Ex. 239. To construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. Ex. 240. To describe circles about the vertices of a given triangle as centres, so that each shall touch the two others. Ex. 241. To bisect the angle formed by two lines, with- out producing the lines to their point of intersection. Draw any line EF II to BA. Take EG = EH, and pro- duce GH to meet BA at I. Draw the J, bisector of GI. -4 F'/ /\f\ ./« /E A C\ yff-'-F B Ex. 242. To draw through a given point P between the sides of an angle BA C a line terminated by the sides of the angle and bisected at P. Ex. 243. Given two points P, Q, and a line AB\ to draw lines fromP and Q which shall meet on AB and make equal angles with AB. Make use of the point which forms with P a pair of points symmetrical with respect to AB. Ex. 244. To find the shortest path from P to Q which shall touch a line AB. Ex. 245. To draw a common tangent to two given circles. Let r and r' denote the radii of the circles, and 0' their centres. With centre and ra- dins r — r' describe a G. ' From (y draw the tan- gents (JM, O'N. Pro- I duce OM and ON to I meet the cu'cumference at A and C. Draw the radii O'B and CD II, respectively, to OA and OC. Draw AB and CD. To draw the internal tangents use an auxiliary O of radius r + 1^. -'-S Book III. PROPORTION. SIMILAR POLYGONS. THE THEORY OF PROPORTION. 323. A proportion is an expression of equality between two equal ratios ; and is written in one of the following forms : a:b = o:d; a:h ■.■.c:d; j = -i' This proportion is read, « a is to 6 as c is to c^ " ; or "the ratio of a to ft is equal to the ratio of c to d." 324. The terms of a proportion are the four quantities com- pared ; the first and third terms are called the antecedents, the second and fourth terms, the consequents ; the first and fourth terms, the extremes, the second and third terms, the means. Thus, in thie proportion a:b = c:d; a and c are tlae antecedents, 6 and d the consequents, a and d the extremes, b and e the means. 325. The fourth proportional to three given quantities is the fourth term of the proportion which has for its first three terms the three given quantities taken in order. Thus, d is the fourth proportional to a, 6, and c in the proportion a : 6 = c : d. 326. The quantities a, h, c, d, e, are said to be in continued proportion, \i a ib = b : c = g : d = d : e. If three quantities are in continued proportion, the second i^ called the mean proportional- between the other two, and the third is called the third proportional to the other two. Tlius, in the proportion a:b = h:c; 6 is tlie mean proportional hetween a and c ; and c is the third proportional to a and 6. 135 136 BOOK III. PLANE GEOMETRY. Proposition I. Theorem. 327. In every proportion the product of the extremes is equal to the product of the means. Let a : b = c : d. Then ^ = 3- § 323 a Whence ad = he. q. e. d. Proposition II. Theorem. 328. The mean proportional between two quantities is equal to the square root of their product. Let a : b = b : c. Then b^ = ae. § 327 Whence, extracting the square root, b = ^. Q. E. D. Ex. 252. If the line joining the middle points of the bases of a trape- zoid is produced, and the two legs are also produced, the three lines will meet in the same point. Ex. 253. AB and AC are chords drawn from any point A in the cir- cumference of a circle, and AD is a diameter. The tangent to the circle at D intersects AB and AC &t E and F, respectively. Show that the triangles ABC and AEF are similar. Ex. 254. AD and BE are two altitudes of the triangle CAB. Show that the triangles CED and GAB are similar. Ex. 255. If two circles are tangent to each other, She chords formed by a straight line drawn through the poin*. of contact have the same ratio as the diameters of the circles. SIMILAR POLYGONS. 157 Proposition XXV. Theorem. 365. If two polygons are similar, they are composed of the same number of triangles, similar each to each, and similarly placed. B C B' Let the polygons ABCDE and A'B'C'D'E' be similar. From two homologous vertices, as U and W, draw diagonals SB, EC, and WB', E'O. To prove that the A EAB, EBC, ECD are similar, respectively, to the A E'A'B', E'B'C, E'C'D'. Proof. The A EAB and E'A'B' are similar. § 357 §361 §361 §351 §351 Ax. 3 §351 §361 Ax. 1 §367 Q.E.D. For /LA = Z.A', and AF,:A'E' = AB:A'B'. Now Z.ABC = ZA'B'C', and •Z ABE = Z A'B'E'. By subtracting, Z EBC = Z E'B'C. Now EB : E'B' = AB : A'B', and BC-.B'C = AB:A'B<. ■.EB:E'B' = BC:B'C'. .-. A EBC and E'B'C are similar. In like manner A ECD and E'C'D' are similar. 158 BOOK III. PLANE GEOMETKY. Proposition XXVI. Theorem. 366. Conversely : If two polygons are composed of the same number of triangles, sim,ilar each to each, and similarly placed, the polygons are similar. In the two polygons ABCDE and A'B'C'D'E', let the triangles AEB, BEC, CED be similar, respectively, to the triangles A'E'B', B'E'C, C'E'D'; and similarly placed. To prove that ABCDE is similar to A'B'C'D'W. Proof. AA = Z. A'. Also, and By adding, AABE = Z.A'B'E\ AEBC = AE'B'C'. §351 §361 Ax. 2 A ABC =Z.A'B'C'. Ill like manner, /.BCD = /. B'C'D', Z CDE = Z CUE', etc, Hence, the polygons are mutually equiangular. Also, -^ - f^^^ _BC _f EC\ CD A'B' \E'B'J B'C ~ \E'C'J ~ In)'' ^^°' ^ ^^^ Hence, the polygons have their homologous sides propor- tional. Therefore, the polygons are similar. §351 Q. E. D. EXERCISES. 159 THEOREMS. Ex. 256. If two circles are taugeut to each other externally, the corre- sponding segments of two lines drawn through the point of contact and terminated by the circumferences are proportional. Ex. 257. In a parallelogram ABCD, a Hue BE is drawn, meeting the diagonal AC in F, the side BC in G, and the side AB produced in F. Prove that JDF' = FG x FE. Ex. 258. Two altitudes of a triangle are inversely proportional to the corresponding bases. Ex. 259. Two circles touch at P. Through P three lines are drawn, meeting one circle ia. A^ B, C, and the other in A', B', C, respectively. Prove that the triangles ABC, A'B'C are similar. Ex. 260. Two chords AB, CD intersect at M, and A is the middle point of the arc CD. Prove that the product AB x AM is constant if the chord AB is made to turn about the fixed point A. Draw the diameter AE, and draw BE. Ex. 261. If two circles touch each other, their common external tan- gent is the mean proportional between their diameters.- Let j4B be the common tangent. Draw tlie diameters AC, BD. Join the poiut of contact P to A, B, C, aud D. Show that APD and BPC are straight lines _L to each other, and that h CAB, ABD are similar. Ex. 262. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Draw any two of the chords, and join the points where they meet the circumferences. The A thus formed are similar (Ex. 120). Ex. 263. In an inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. Draw DE, making Z CDE = AADB. The A ABD and ECD are similar ; and the A BCD and AED are similar. . Ex. 264. Two isosceles triangles with equal vertical angles are similar. Ex. 265. The bisector of the vertical angle A of the triangle ^BC inter- sects the base at D and the circumference of the circumscribed circle at E. Show that AB X AC = AD x AE. 160 BOOK III. PLANE GEOMETRY. NUMERICAL PROPERTIES OF FIGURES. Proposition XXVII. Theorem.. 367. If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse : 1. The triangles thus formed are similar to the given triangle, and to each other. 2. The perpendicular is the m,ean proportional hetween the segments of the hypotenuse. 3. Each leg of the right triangle is the mean propor- tional between the hypotenuse and its adjacent segm,ent. In the right triangle ABC, let CP be drawn from the vertex of the right angle C, perpendicular to AB. 1. To prove that A BCA, CFA, BFC are similar. Proof. The rt. A CFA and BCA are similar, § 356 since the Z. a' is common. The rt. A BFC and BCA are similar, § 356 since the Z.b is common. Since the A CFA and BFC are each similar to A BCA, they are similar to each other. § 354 2. To prove that AF : CF = CF : FB. Proof. In the similar A CFA and BFC, AF : CF = CF : FB. § 351 NUMERICAL PROPERTIES OF FIGURES. 161 3. To prove that AB:AC = AC: AF, and AB:BC = BC:BF. Proof. In the similar A BCA and CFA, AB : AC = AC : AF. §351 In the similai- A BCA and BFC, AB:BC = BC:BF. § 351 Q.E.D. 368. CoE. 1. The squares of the two legs of a right tri- angle are proportional to the adjacent segments of the hypote- nuse. From the proportions in § 367, 3, AA? = ABX AF, and BC^ = AB x BF. § 327 7C' AB X AF AF Hence, ==; = = BC AB XBF BF 369. Cor. 2. The squares of the hypotenuse and either leg are proportional to the hypotenuse and the adjacent segment. Iff AB X AB AB Eor AC'' AB X AF AF 370. CoE. 3. The perpendicular from any point in the circumference to the diameter of a circle is the mean proportional between the seg- ments of the diameter. The chord drawn from any point in the circumference to either extremity of the diameter is the mean proportional between the diameter and the adjacent segment. For the ZACBisa. rt. Z. § 290 162 BOOK III. PLANE GEOMETRY. Proposition XXVIII. Theorem. 371. The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. Let ABC be a right triangle with its right angle at C. To prove that AC' + CB^ = AB^. Proof. Draw OF J. to AB. Then and AC =^ABX AF, CB'' = ABx BF. §367 Q.E.D. Ey adding, AC' + CB' = AB{AF + BF) = AB 372. Cor. 1. The square of either leg of a right triangle is equal to the difference of the square of the hypotenuse and the square of the other leg. 373. Cor. 2. The diagonal and a side of a square are incommensurable. For JC' = 'aF + 'BC'' = 2AB\ .■.ac = abV2. 374. Def. The projection of any line upon a second line is the segment of the second line included between the perpendiculars drawn to it from the extremities of the first line. Thus, A.- PB is the projection of CD upon AB. NUMERICAL PROPERTIES OE FIGURES. 163 Proposition XXIX. Theoeem. 375. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished hy twice the product of one of those sides by the projection of the other upon that side. Let C be an acute angle of the triangle ABC, and DC the projec- tion of AC upon BC. To prove that AB" = 'bG^ + Ic' -2BC X BC. '' Proof. If D falls upon tlie base (TFig. 1), DB = BC - BC. If D. falls upon the base produced (Fig. 2), BB = BC - BC. In either case, 'bF = l3G^ + 1)G^ - 2 BC XBC. Add AB to both sides of this equality, and we have AB' + BB' = BC'' + AB' + BC'-2BC XBC. But AB' + BB' =: AB\ and Ib" + BC^ = AC''. § 371 Put AB and A C for their equals in the above equality. Then AB^ = BC' + AC'-2BC X BC. q.e.d. 164 BOOK III. PLANE GEOMETRY. Proposition XXX. Theokem. 376. In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides by the projection of the other upon that side. Let C be the obtuse angle of the triangle ABC, and CD be the pro- jection of AC upon BC produced. To prove that AB" = BC^ + AC' + 2BC X DC. Proof. DB = BC + DC. Squaring, IdF = BC' + Wf- + 2BCy.DC. Add ad' to both sides, and we have aT? + In? = Tcf- + Zd' + 'DC? + 2BC x-dc. But ad' + DB^ = AB\ and AD' + Ixf = 2c'- § 371 Put AB' and A C for their equals in the above equality. Then Ib' = 'BC^ + AG^+2BCx DC. q.e.d. Note 1. By the Principle of Continuity the last three theorems may he included in one theorem. Let the student explain. Note 2. The last three theorems enable us to compute the lengths of the altitudes of a triangle if the lengths of the three sides axe known. NUMERICAL PROPERTIES OP FIGURES. 165 Pkoposition XXXI. Theorem. 377. 1. TTie sum of the squares of two sides of a tri- angle is equal to twice the square of half the third side increased by twice the square of the median upon that side. 2. The difference of the squares of two sides of a tri- angle is equal to twice the product of the third side by the projection of the median upon that side. In the triangle ABC, let AM be the median and MD the projection of AM upon the side BC. Also, let AB be greater than AC. To prove that 1. Zb' + AC" = 2 BM' + 2 1m\ 2. 1^-20^ = 230 X MD. Proof. Since AB >A0, the Z AMB will be obtuse, and the Z AMO will be acute. § 155 Then AB" = 'bM^ + 'AM^ + 2 BM X MD, § 376 and IC^ = MC'' + Zm"' - 2 JfC X MD. § 375 Add these two equalities, and observe that BM = MC. Then 'AB' + AC'' = 2BM^ + 2AM\ Subtract the second equality from the first. Then IF - AC'' = 2BC X MD. q.e.d. Note. This theorem enables us to compute the lengths of the medians of a triangle if the lengths of the three sides are known. 166- BOOK III. PLANE GEOMETRY. Proposition XXXII. Theorem. 378. If two chords intersect in a circle, the product of ike segments of one is equal to the product of the segments of the other. Let any two chords MN and PQ intersect at 0. To prove that OM X 0N= OQ X OP. Proof. Draw MP and NQ. Za = Za', § 289 (each being measured by-\ arc FN). And /^c = /.c\ § 289 (each being measured by -J- arc MQ). .-.the A WO Q and POM are similar. § 355 .■.OQ:OM=ON:OP. §351 .-. OMxON=OQxOP. §327 Q.E.D. 379. Scholium. This proportion may be written OM_OP^ OM _ 1 OQ ~ on' ""^ OQ ~'qN' OP that is, the ratio of two corresponding segments is equal to the reciprocal of the ratio of the other two segments. Hence, these segments are said to be reciprocally proportional NUMERICAL PROPERTIES OF FIGURES. 167 380. Def. a secant from a point to a circle is understood, to meaa the segment of the secant lying between the point and the second point of intersection of the secant and circumference. Pkoposition XXXIII. Theorem. 381. If from a point loithout a circle a secant and a tangent are drawn, the tangent is the mean proportional between the whole secant and its external segment. Let AD be a tangent and AC a secant drawn from the point A to the circle BCD. To prove that AC : AD = AD: AJB. Proof. Draw DC and DB. The A ADC and ADD are similar. § 355 Por Z S is common ; and Z a' = Z a, §§ 289, 295 (each being measured by i arc BB). .■ . AC : AD = AD : AB. §351 Q.E.D. 382. CoE. If from a fixed point without a circle a secant is drawn, the product of the secant and its external segment is constant in whatever direction the secant is drawn. For ACxAB = AD\ § 327 168 BOOK III. PLANE GEOMETRT. Proposition XXXIV. Theorem. 383. The square of the bisector of an angle of a tri- angle is equal to the product of the sides of this angle diminished hy the product of the segments made by the bisector upon the third side of the triangle. Let WO bisect the angle MNP of the triangle MNP. To prove that 'Wcf = NM X NP - OM X OP. Proof. Circumscribe the O MNP about the A MNP. § 314 Produce NO to meet the circumference in Q, and dra^w PQ. The ANQP and NMO are similar. § 355 Hyp. §289 §351 For Zb=Ab', and Za = Za'. Whence NQ : NM = NP : NO. .-. NM XNP = NQ X NO = (NO+OQ)NO = NO' + NO X OQ. But NO X OQ = MO X OP. .-. MN XNP = no" + MOx op. Whence NO^ = NM X NP - MO X OP. § 378 Ax. 3 Q. E. D. Note. This theorem enables us to compute the lengths of the bisectors of ttie angles of a triangle if the lengths of the sides are known. NUMERICAL PROPERTIES OF FIGURES. 169 Proposition XXXV. Thboeem. 384. In any triangle the product of two sides is equal to the product of the diameter of the circumscribed circle by the altitude upon the third side. Let WMQ be a triangle, NO the altitude, and QWMP the circle cir- cumscribed about the triangle NMQ. Draw the diameter iVP, and draw PQ. To prove that NM X NQ = NP X NO. Proof. In the A NOM and NQP, Z NOM is a rt. Z, Hyp. Z NQP is a rt. Z, § 290 and Z a = Z a', § 289 (each being measured by ^ arc NQ). .-.A NOM and NQP are similar. § 356 Whence NM ■.NP = NO: NQ. § 361 .-. NM X NQ = NP X NO. § 327 Q.E.D. Note. This theorem enables us to compute the length of the radius of a circle circumscribed about a triangle, if the lengths of the three sides of the triangle are known. Ex. 266. If OE, OF, OG are the perpendiculars from any point with in th e triangle AB C u pon the sides AB, BC, OA, respectively, show that Ae'^ + BF^ + CG^ = F^ + FC^ + GA\ 170 BOOK III. PLANE GEOMETRY. THEOREMS. Ex. 267. The sum of the squares of the segments of two perpendicular chords is equal to the square of the diameter of the circle. If AB, CD are the chords, draw the diameter BE, draw .4 C, ED, BD. Prove that AG = ED, and apply § 371. Ex. 268. The tangents to two intersecting cii-cles drawn from any point In their common chord produced, are equal. (§ 381.) Ex. 269. The common chord of two intersecting circles, if produced, will bisect their common tangents. (§ 381.) Ex. 270. If thi'ee circles intersect one another, the common chords all pass through the same point. Let two of the chords AB and CD meet at 0. Join - the point of intersection E to 0, and suppose that EO produced meets the same two circles at two different points P and Q. Then prove that OP = OQ (§ 378); hence, that the points P and Q coincide. Ex. 271. If two circles are tangent to each other, the common internal tangent bisects the two common external tangents. Ex. 272. If the perpendiculars from the vertices of the triangle ABC upon the opposite sides intersect at D, show that aW - AC^ = BD^ - 'GD'. Ex. 273. In an isosceles triangle, the square of a leg is equal to the square of any line drawn from the vertex to the base, increased by the product of the segments of the base. Ex. 274. The squares of two chords drawn from the same point in a circumference have the same ratio as the projections of the chords on the diameter drawn from the same point. Ex. 275. The difference of the squares of two sides of a triangle is equal to the difference of the squares of the segments of the third side, made by the perpendicular on the third side from the opposite vertex. Ex. 276. E is the middle point of BC, one of the parallel sides of the trapezoid ABCD; AE and DE produced meet DC and AB produced at F and O, respectively. Show that FG is parallel to DA. AAGD and BGE are similar ; and A AFD and EFG are similar, EXERCISES. 171 Ex. 277. If two tangents are drawn to a circle at the extremities of a diameter, the portion of a third tangent intercepted between them is divided at its point of contact into segments whose product is equal to the squai'e of the radius. Ex. 278. If two exterior angles of a triangle are bisected, the line drawn from the point of intersection of the bisectors to the opposite angle of the triangle bisects that angle. Ex. 279. The sum of the squares of the diagonals of a quadrilateral is equal to twice the sum of the squares of the lines that join the middle points of the opposite sides. Ex. 280. The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of the diagonals, in- creased by four times the square of the line jo'iniug the middle points of the diagonals. Apply § 377J0 the A ABC and ABC, add the results, and eliminate BJS^ + Dif by applying § 377 to the A BDS. Ex. 281. The square of the bisector of an exterior angle of a triangle is equal to the product of the external segments determined by the bisector upon one of the sides, diminished by the ?;_-. product of the other two sides. Let CD bisect the exterior / BCH of the A ABC. A ^ CD and FCB are similar (§ 355). Apply § 382. Ex. 282. If a point is joined to the vertices of a triangle ABC ; through any point A' in OA a line parallel to AB is drawn, meeting OB at B' ; through B' a line parallel to BC, meeting OC at C ; and C is joined to A' ; the triangle A'B'C is similar to the triangle ABC. Ex. 283. If the line of centres of two circles meets the circiunferences at the consecutive points A, B, C, D, and meets the common external tan- gent at P, then PA x PD = PB x PC. Ex. 284. The line of centres of two circles meets the common external tangent at P, and a secant is drawn fi-om P, cutting the circles at the consecutive points E, F, G, H. Prove that PE x PH = PF x PG. Draw radii to the points of contact, and to E, F, G, H. Let fall Js on PS from the centres of the ®. The various pairs of A are similar. Ex. 285. If a line drawn from a vertex of a triangle divides the oppo-- site side into segments proportional to the adjacent sides, the line bisects the angle at the vertex. 172 BOOK III. PLANE GEOMETRY. PROBLEMS OF COWSTKUCTION. Proposition XXXVI. Problem. 385. To divide a given straight line into parts pro- portional to any number of given lines. H K B Let AB, m, n, and p, be given straight lines. To divide AB into parts proportional to m, n, and p. Draw AX, making any convenient Z. with AB. On AX take AC equal to m, CE to n, EF top. Draw BF. From E and C draw EK and CII II to FB. Tlirough A draw a line II to BF. K and H are the division points required. AH HK KB ^•"'*- AG=^ = EF' §^^ (if two lines are cut hy any number of parallels, the corresponding inter- cepts are proportional). Substitute m, n, and p for their equals A C, CE, and EF. AH HK KB Then = ■ = m n p Q EF Ex. 286. Divide a line 12 inches long into three parts proportional to the numbers 3, 5, 7. PROBLEMS OF CONSTRUCTION. 173 Pboposition XXXVII. Problem. 386. To find the fourth proportional to three given straight lines. -. Let the three given lines be m, n, and p. To find the fourth proportional to m, n, and p. Draw Ax and Ay containing any convenient angle. On Ax take AlB equal to m, BC to n. On Aij take AD equal to p. Draw BD. From C draw CF II to BD, meeting Ay at F. DF is the fourth proportional required. Proof. AB-.BC = AD: DF, § 342 (a line drawn through two sides of a A II to the third side divides those sides proportionally) . Substitute m, n, and^ for their equals AB, BC, and AD. Then m:n = p: DF. q. e. f. Ex. 287. The square of the altitude of an equilateral triangle is equal to three fourths of the square of one Side of the triangle. 174 BOOK III. PLANE GEOMETRY. Proposition XXXVIII. Peoblbm. 387. To find the third proportional to two given straight lines. IB/-— -\p jjL \E Let m and n be the two given straight lines. To find the third proportional to m and n. Construct any convenient angle A, and take AB equal to m, ^C equal to n. Produce AB to D, making BD equal to A G. Draw BC. Through D draw DE II to BC, meeting AC produced at E. CE is the third proportional required. Proof. AB : BD = AC : CE, § 342 {a line drawn through two sides of a A parallel to the third side divides those sides proportionally). Substitute, in the above proportion, A C for its equal BD. Then AB : AC = AC : CE, that is, m-.n = n: CE. o-E.f. Ex. 287. Construct x, if (1) x = — , (2) a; = -• c c Special oases : (1) a = 2, 6 = 3, c = 4 ; (2) a = 3, 6 = 7, c = 11 ; (3) a = 2, c = 3 ; (4) a = 3, c = 5 ; (5) a = 2 c. PROBLEMS OF CONSTEXJCTION. 175 Proposition- XXXIX. Problem. 388. To find the mean proportional between two given straight lines. A L .; _iT m c n B Let the two given lines be m and n. To find the mean proportional between m and n. On the straight line AE take AC equal to m, and CB equal to n. On AB as a diameter describe a semicircumference. At C erect the X CH meeting the circumference at H. CM is the mean proportional between m and n. Proof. AC:CH=CH: CB, § 370 {the J, Id fall from a point in a circumference to the diameter of a circle is the mean proportional between the segments of the diameter). Substitute for AG and CB their equals m and n. Then m:CH= CH: n. q.e f. 389. Def. a straight line is divided in extreme and mean ratio, when one of the segments is the mean proportional between the whole line and the other segment. Ex. 289. Construct x,iix = Va6. Special cases : (1) a = 2, 6 = 3 ; (2) a = 1, 6 = 5 ; (3) a = 3, 6 = 7. 176 BOOK III. PLANE GEOMETRY. Proposition XL. Problem. 390. To divide a given line in extreme and mean ratio. E,.- O^- I / A C B Let AB be the given line. To divide AB in extreme and mean ratio. At B erect a ± BE equal to half of AB. From j& as a centre, with a radius equal to EB, describe a O. Draw AJS, meeting the circumference in F and G. On AB take AC equal to AF. On BA produced take AC equal to AG. Then AB is divided internally at C and externally at C in extreme and mean ratio. Proof. AG:AB AB'' = AFx AG = AC(AF + FG) = AC{AC + AB) = AC'' + ABx AC. .-.ab'-abxac^ac"^ .•.ab(ab-ac) = ac\ .-.AB X CB = 'AC^. ■■ AB : AF. §381 = AGxAF = C'A{AG - FG) = C'A{C'A-AB) = 'CAl -ABx C'A. Aff + ABx C'A = 1Ja\ ■.AB{AB + CA) = 'CA\ AB X C'B = 'CA\ Q.E.F. PROBLEMS OF CONSTRUCTION. 177 Peoposition XLI. Peoblbm. 391. Upon a given line homologous to a given side of a given polygon, to construct a polygon similar to the given polygon. If I B' a' Let A'E' be the given line homologous to AE of the given polygon ABCDE. To construct on A'H' a polygon similar to the given polygon. rrom E draw the diagonals EB and EC. From E' draw E'B', E'C, and £"/)', vasikingAA'E'B', B'E'C, and C'E'D' equal, respectively, to A AEB, BEG, and CED. From A' draw A'B', making Z EA'B' equal to Z EAB, and meeting E'B' at B'. From B' draw B'C, making Z E'B'C equal to Z EBC, and meeting E'C at C". From C" draw C'i?', making Z E'C'D' equal to Z ^C2) and meeting E'D' at D'. Then A'B' CD' E' is the required polygon. Proof. The A ^5^, A'B'E', etc., are similar. § 354 Therefore, the two polygons are similar, § 366 178 BOOK III. PLANE GEOMETRY. PROBLEMS OP CONSTEUCTION. Ex. 290. To divide one side of a given triangle into segments propor- tional to tlie adjacent sides (§ 318). Ex. 291. To find in one side of a given triangle a point whose distances from the other sides shall be to each other in the given ratio m : n. Take ^G = m J, to AC, GH=nX to BC. Draw CD II t« OG. Ex. 292. Given an obtuse triangle ; to draw a line from the vertex of the obtuse angle to the opposite side which shall be the mean proportional between the segments of that side. Ex. 293. Through a given point P within a given circle to draw a chord AB so that the ratio AP : BP shall equal the given ratio m ; n. Draw OPC so that OP : PC = n:m. Draw CA equal to the fourth proportional to n, m, and the radius of the circle. Ex. 294. To draw through a given point P in the arc subtended by a chord AB a chord which shall be bisected by AB. On radius OP take CD equal to CP. Draw DE II to BA. Ex. 295. To draw through a given external point P a secant PAB to a given circle so that the ratio PA : AB shall equal the given ratio m : n. PD:DC = m: n. PD:PA = PA: PC. Ex. 296. To draw through a given external point P a secant PAB to a given circle so that Al^ = PA x PB. PC: CD = CD: PD. PA = CD. EXERCISES. 179 Ex. 297. To find a point P in the arc subtended by a given chord AB so that the ratio PA : PB shall equal the given ratio m : n. Ex. 298. To draw tlu'ough one of the points of intersection of two ch'cles a secant so that the two chords that ai'e formed shall be in the given ratio m : n. Ex. 299. Having given the greater segment of a Une divided in extreme and mean ratio, to construct the line. Ex. 300. To construct a circle which shall pass tlu'ough two given points and touch a given straight line. Ex. 301. To construct a circle which shall pass through a given point and touch two given straight lines. Ex. 302. To inscribe a square in a semicircle. Ex. 303. To inscribe a square in a given triangle. Let DEFG be the required inscribed square. Draw CM II to AB, meet- ing AF produced in M. Draw CU and MN Xto AB, and produce AB to meet MN at N. The A ACM, AGF are ?- --¥ similar; also, the A AMN, AFE are similar. By these al^-''' i triangles show that the iigure CMNH is a, square. By ^-- ' |\ i constructing this square, the point F can be found. ^ vueb n Ex. 304. To inscribe in a given triangle a rectangle similar to a given rectangle. Ex. 305. To inscribe in a circle a triangle similar to a given triangle. Ex. 306. To inscribe in a given semicircle a rectangle similar to a given rectangle. Ex. 307. To circumscribe about a circle a triangle similar to a given triangle. 2abc ^, ^ . 2 db c Ex. 308. To construct the expression, x = -^ ; that is, —^ x -• Ex. 309. To construct two straight lines, having given their sum and their ratio. Ex. 310. To construct two straight liues, havmg given their difference and their ratio. Ex. 311. Given two circles, with centres and C, and a pomt A in thei^ plajie, to draw through the point A a straight Ime, meeting the ou- cumferences at B and C, so that AB ■.AC = m:n. 180 BOOK III. PLANE GEOMETRY. PROBLEMS OF COMPUTATION. Ex. 312. To compute the altitudes of a triangle in terms of its sides. a B c A D At least one of the angles ^ or B is acute. Suppose B is acute. In the A CDB, h? = a^ - B&. § 372 In the A ABC, 62 = a^ + c^ - 2 c x BZ). § 375 a2 + c2 - 62 Whence Hence, BB: 2c Let Then Hence, 2 (02 -[■ e2 - 62)2 ^ 4 g2fa _ (^2 + e2 - 62)2 _ (2 ae + a2 + c2 - 62) (2 ac - 0,"^ - c^ + 62) ~ 4c2 ' _ {(a + c)2 - 62} {52 - (g - c)2} ~ 4c2 _ (g + 6 + e) (g + c - 6) (5 + g - c) (6 - g + c) ~ 4c2 a + 6 + c = 2s. g + c -6 = 2(8-6), 6 + a — c = 2 (s — c), 6 — g + c = 2 (s — g). _ 2 s X 2 ( s - g) X 2 (s - 6) X 2 (s - c) '^ - rE2 By simplifying, and extracting the square root, 2 I A = - Vs (s - g) (s - 6)(s - c). Ex. 313. To compute the medians of a triangle in terms of its sides. By §377, g2 + 62 = 2m2 + 2(|y. ^ Whence 4 )?i2 = 2 (o2 + 62) - c2. .-. m = i V2 (g2 + 62) _ c2. EXERCISES. 181 Ex. 314. To compute the bisectors of a triangle in terms of the sides. By § 383, t:^ = ab-ADx BB. 6 a a + b .: AD = be , and BD -. Whence a + b ~o + & ac a + b' ahe^ Whence « = {a + 6)2 _ ab{{a + bY-c^} (a + 6)2 _ a6 (g + 5 + c) (g + 5 — e) (a + 6)2 _ a5 X 2 s X 2 (« — c) ~ (a + 6)2 2 g + 6 -^abs{s — c). Ex. 315. To compute the radius of the circle circumscrihed about a tri- angle in terms of the sides of the triangle. By § 384, AC xAB = AE X AD, or, 6c = 2 iJ X AD. 2 / -B/- But AD = - Vs(s-a)(s-6){s-c). Ex. 312 .B = . dbc 4 Vs (s — a) (s — 6) (s — c) Ex. 316. If the sides of a triangle are 3, 4, and 5, is the angle opposite 5 right, acute, or obtuse ? Ex. 317. If the sides of a triangle are 7, 9, and 12, is the angle opposite 12 right, acute, or obtuse ? Ex. 318. If the sides of a triangle are 7, 9, and 11, is the angle opposite 11 right, acute, or obtuse ? Ex. 319. The legs of a right triangle are 8 inches and 12 inches ; find the lengths of the projections of these legs upon the hypotenuse, and the distance of the vertex of the right angle from the hypotenuse. Ex. 320. If the sides of a triangle are 6 Inches, 9 inches, and 12 inches, find the lengths (1) of the altitudes ; (2) of the medians ; (3) of the bisec- tors; (4) of the radius of the circumscribed circle. 182 BOOK III. PLANE GEOMETRY. Ex. 321. A line is drawn parallel to a side AB of a triangle ABC, cutting ^0 in D, BC in E. If AD : DC = 2:3, and AB = 20 inches, find DE. Ex. 322. The sides of a triangle are 9, 12, 15. Find the segments of the sides made by bisecting the angles. Ex. 323. A tree casts a shadow 90 feet long, when a post 6 feet high casts a shadow 4 feet long. How high is the tree ? Ex. 324. The lower and upper bases of a trapezoid are a, 6, respec- tively ; and the altitude is h. Find the altitudes of the two triangles formed by producuig the legs until they meet. Ex. 325. The sides of a triangle are 6, 7, 8, respectively. In a similar triangle the side homologous to 8 is 40. Find the other two sides. Ex. 326. The perimeters of two similar polygons are 200 feet and 300 feet. If a side of the first is 24 feet, find the homologous side of the second. Ex. 327. How long a ladder is required to reach a window 24 feet high, if the lower end of the ladder is 10 feet from the side of the house ? Ex. 328. If the side of an equilateral triangle is a, find the altitude. Ex. 329. If the altitude of an equilateral triangle is h, find the side. ' Ex. 330. Find the length of the longest chord and of the shortest chord that can be drawn through a point 6 inches from the centre of a circle whose radius is 10 inches. Ex. 331. The distance from the centre of a circle to a chord 10 feet long is 12 feet. Find the distance from the centre to a chord 24 feet long. Ex. 332. The radius of a circle is 5 inches. Through a point 3 inches from the centre a diameter is drawn, and also a chord perpendicular to the diameter. Find the length of this chord, and the distance from one end of the chord to the ends of the diameter. Ex. 333. The radius of a circle is 6 inches. Find the lengths of the tangents drawn from a point 10 inches from the centre, and also the length of the chord joining the points of contact. Ex. 334. The sides of a triangle are 407 feet, 368 feet, and 351 feet Find the three bisectors and the thi-ee altitudes. EXERCISES. 183 Ex. 335. If a chord 8 inches long is 3 inches distant from the centre of the circle, find the radius, and the chords drawn from the end of the choi-d to the ends of the diameter which bisects the chord. Ex. 336. From the end of a tangent 20 inches long a secant is drawn through the centre of the circle. If the external segment of this secant is 8 inches, find the radius of the circle. Ex. 337. The radius of a circle is 13 inches. Through a point 5 inches from the centre any chord is drawn. "What is the product of the two segments of the chord ? What is the length of the shortest chord that can be drawn through the point ? Ex. 338. The radius of a circle is 9 inches and the length of a tangent 12 inches. Find the length of a line drawn from the extremity of the tangent to the centre of the circle. Ex. 339. Two cu-cles have radii of 8 inches and 3 inches, respectively, and the distance between their centres is 15 inches. Find the lengths of their common tangents. Ex. 340. Find the segments of a line 10 inches long divided in extreme and mean ratio. Ex. 341. The sides of a triangle are 4, 5, 5. Is the largest angle acute, right, or obtuse ? Ex. 342. Find the third proportional to two lines whose lengths are 28 feet and 42 feet. Ex. 343. If the sides of a triangle are u., 6, 6, respectively, find the lengths of the three altitudes. Ex. 344. The diameter of a cu-cle is 30 feet and is divided into five equal parts. Find the lengths of the chords drawn through the points of division perpendicular to the diameter. Ex. 345. The radius of a ciixle is 2 inches. From a point 4 inches from the centre a secant is drawn so that the internal segment is 1 inch. Find the length of the secant. Ex. 346. The sides of a triangular pasture are 1551 yards, 2068 yards, 2585 yards. Find the median to the longest side. Ex. 347. The diagonal of a rectangle is d!, and the perimeter is p. Find the sides. Ex. 348. The radius of a circle is r. Find the length of a chord whose distance from the centre is ^r. Book IV. AEEAS OF POLYGONS. 392. Def. The unit of surface is a square "wiiose side is a unit of length. 393. Def. The area of a surface is the number of units of surface it contains. 394. Def. Plane figures that have equal areas but cannot be made to coincide are called equivalent. Note. In propositions relating to areas, tlie words "rectangle," " tri- angle, " etc. , are often used for ' ' area of rectangle, " " area of triangle, ' ' etc. Peoposition I. Theoebm. 395. Two rectangles having equal altitudes are to each other as their bases. o Let the rectangles AC and AF have the same altitude AD. To prove that rect. AC : rect. AF = base AB : base AM Case 1. When AB and AJE are commensurable. Proof. Suppose AB and AE have a common measure, as AO, which is contained m times in AB and n times in AE. Then AB:AE==m:n. 184 AREAS OP POLYGONS. 185 Apply AO as a, unit of measure to A£ and AE, and at the several points of division erect Js. The rect. ^C is divided into m rectangles, and the rect. AF is divided into n rectangles. § 107 These rectangles axe all equal. § 186 Hence, rect. A : rect. AF = m:n. Therefore, rect. A C : rect. AF.= AB : AK Ax. 1 Case 2. When AB and AJS are incommensurable. P H F K JE Proof. Divide AB into any number of equal parts, and apply one of them to AE as many times as AF will contain it. Since AB and AF are incommensurable, a certain number of these parts will extend from A to some point K, leaving a remainder KF less than one of the equal parts of AB. Draw KH II to FF. Then AB and AK are commensurable by construction. „, „ rect. AIT AK ^ ^ Therefore, — ^'77f = '7^" Case 1 ' rect. AC AB If the number of equal parts into which AB is divided is indefinitely increased, the varying values of these ratios will continue equal, and approach for their respective limits the ratios ^52*141 and 41- (See §287.) rect. AC AB ^ rect. AF AF AB §284 Q.E.D. rect. A C 396. CoE. Two rectangles having equal bases are to each other as their altitudes. 186 BOOK IV. PLANE GEOMETRY. Proposition II. Theorem. 397. Two rectangles are to each other as the products of their bases by their altitudes. R' Let R and R' be two rectangles, having for their bases b and b', and for their altitudes a and a', respectively. ^_ g X 5 R ~ To prove that a' X b' Proof. Construct the rectangle S, with its base equal to that of R, and its altitude equal to that of R'. The products of the corresponding members of these equa- tionsgive ig_ axb R' Then and a' X b' Q.E.D. Ex. 349. rind the ratio of a rectangular lawn 72 yards ty 49 yards to a grass turf 18 inches by 14 inches. Ex. 350. Find the ratio of a rectangular courtyard 18^ yards by 15J yards to a flagstone 31 inches by 18 inches. Ex. 351. A square and a rectangle have the same perimeter, 100 yards. The length of the rectangle is 4 times its breadth. Compare their areas. Ex. 352. On a certain map the linear scale is 1 inch to 5 miles. How many acres are represented on this map by a square the perimeter of which is 1 inch ? AREAS OF POLYGONS. 18T Proposition III. Theorem. 398. The area of a rectangle is equal to the product of its base by its altitude. ■0 Let R be a rectangle, b its base, and a its altitude. To prove that the area of R ^ a Y. b. Proof. Let U be the unit of surface. R _ a Xb Z7~l X 1^ a X 6, 397 (f,wo rectangles are to each other as the products of their bases and altitudes). But — = the number of units of surface in R. .". the area oi R = a X b. 393 (J.E.D. 399, Scholium. When the base and altitude each contain the linear unit an integral number of times, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of surface. Thus, if the base contains seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of surface. 188 BOOK IV. PLANE GBOMETEY. Pkoposition IV. Theorem. 400. The area of a parallelogram is equal to the product of its base by its altitude. A b D Let A£FD be a parallelogram, b its base, and a its altitude. To prove that the area of the O AEFD — a X. b. Proof. Prom A draw AS II to Z>C to meet FE produced. Then the figure ABCD is a rectangle, with the same base and the same altitude as the O AEFD. § 167 The rt. A ABE and DCF are equal. § 151 For AB = CD, and AE = DF. § 178 From ABED take the A DCF; the rect. ABCD is left. From ABED take the A ABE; the O AEFD is left. .-. rect. ABCD^n AEFD. Ax. 3 But the area of the rect. ABCD = a X b. § 398 .-.the area of the O AEFD = a X b. Ax. 1 Q.E.D. 401. Cob. 1. Parallelograms having equal bases and equal altitudes are equivalent. 402. CoE. 2. Parallelograms having equal bases are to each other as their altitudes ; parallelograms having equal alti- tudes are io each other as their bases; any two parallelo- grams are to each other as the products of their bases by their altitudes. AEEAS OF POLYGONS. 189 Proposition V. Theorem. 403. The area of a triangle is equal to half the product of its base by its altitude. A b B Let a be the altitude and b the base of the triangle ABC. To prove that the area of the A AJBC — ^a X b. Proof. Construct on AB and BC the parallelogram ABCH. Then A ABC = i O ABCH. § 179 The area oi the EJ ABCIT = a X b. § 400 Therefore, the area of A ABC = ^a X b. Ax. 7 Q.E.D. 404. CoK. 1. Triangles having equal bases and equal alti- tudes are equivalent. 405. Cor. 2. Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the products of their bases by their altitudes. 406. Cor. 3. The product of the legs of a right triangle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle. Ex. 353. The lines which join the middle point of either diagonal of a quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts. 190 BOOK IV. PLANE GEOMETRY. Proposition VI. Theorem. 407. The area of a trapezoid is equal to half the sum of its bases multiplied by the altitude. H Eh' q 1 X A J \p 1 f 1 y a \ A F h B Let b and b' be the bases and a the altitude of the trapezoid ABCH. To prove that the area of ABCH = ^a(b + b'). Proof. Draw the diagonal AC. Then the area of the A J5C = i a. X 5, and the area of the A AHC = ^aXh'. § 403 .-.-the areaof ^J?C^ = ia(6 + 6'). Ax. 2 Q.E.D. 408. Cor. The area of a trapezoid is equal to the product of the median hy the altitude. § 190 409. Scholium. The area of an irregular polygon may be found by dividing the poly- gon into triangles, and by find- ing the area of each of these triangles separately. Or, we may draw the longest diago- nal, and let fall perpendiculars upon this diagonal from the other vertices of the polygon. The sum of the areas of the right triangles, rectangles, and trapezoids thus formed is the area of the polygon. AREAS OF POLYGONS. 191 Proposition VII. Theorem. 410. The areas of tioo triangles ivhich liaue an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A Let the triangles ABC and ADE have the common angle A. A ABC ABxAC To prove that Proof. Now A ADi; ABX AH Draw JBK and A ABC A ABU A ABU AC AJS' AB AD § 405 A ADE The products of the first members and of the second members of these equalities give A ABC _ ABx AC ' ^ ^' " Q.E.D. A ADE AD X AE Ex. 354. The areas 'of two triangles which have an angle of tlie one supplementary to an angle of the other are to each other as the products of the sides including the supplementary angles. 192 BOOK IV. PLANE GEOMETRY. COMPARISON OF POLYGONS. Proposition VIII. Theorem. 411. The areas of two similar triangles are to each other as the squares of ahy two homologous sides. Let the two similar triangles be ACB and A'C3'. AACB IF To prove that = „ • AA'C'B' A'B^ Proof. Draw the altitudes CO and CO'. Then AACB ABx CO AB ^ y CO §405 A A'C'B' A'B' X CO' A'B' CO' (two ^ are to each other as the products of their 6ases by their altitudes). AB CO But A'B' CO' §361 (iAe homologous altitudes of two. similar A have the same ratio as any two homologous sides). CO . . AB Substitute, in the above equality, for "t^ttt, its equal -jr^, j then AB AB AB' X AACB AA'C'B' A'B' A'B' A'B'' Q.E.D Ex. 355. Prove this proposition by § 410. COMPARISON OF POLYGONS. 193 Proposition IX. Theorem. 412. The areas of two similar polygons are to each other as the squares of any two homologous sides. Let S and S' denote the areas of the two simUar polygons ABC etc. and A'B'C etc. To prove that S : S' = AB" : A'B'^. Proof. By drawing all the diagonals from any homologous vertices E and E', the two similar polygons are divided into similar triangles. § 366 A ABE fBE\ ABGE , „ ,_ = = ( =■. \ = = etc. § 411 AA'B'E' \B'E'J AB'C'E' A'B'' A ABE ABCE A ODE That is, AA'B'E' AB'C'E' ' AC'D'E' AABE + ABCE + ACDE A ABE AF A A'B'E' + AB'C'E' + A C'D'E A A'B'E' A'B'' .-. S-.S'^IF-.A^'^ 335 Q.E.D. 413. Cor. 1. The areas of two similar polygons are to each other as the squares of any two homologous lines. 414. Cor. 2. The homologous sides of two similar polygons have the same ratio as the square roots of their areas. 194 BOOK IV. PLANE GEOMETRY. Peoposition X. Theorem:. 415. The square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the two legs. Let BE, CH, AF be squares on the three sides of the right tri- angle ABC. To prove that BE =s= CH + AF. Proof. Through A draw AL II to CE, and draw AD and CF. Since ABAC, BAG, and CAH are rt. A, CA G and BAH are straight lines. §90 The AABD = AFBC. § 143 Por BD = BC, BA = BF, §168 and AABD = ZFBC, Ax. 2 L E (each being the sum of art. Z and the ZABG). Now the rectangle BL is double the A ABD, (ha,ving the same base BB, and the same altitude, the distance between the lis AL and BD), and the square AF is double the A FBC, (having the same base FB, and the same altitude AB). -•. the rectangle BL is equivalent to the square AF. Ax. 6 In like manner, by drawing AF and BK, it may be proved that the rectangle CL is equivalent to the square CH. Hence, the square BF, the sum of the rectangles BL and CL, is equivalent to the sum of the squares C^and AF. q.e.d. 416. Cor. The square on either leg of a right triangle is equivalent to the difference of the square on the hypotenuse and the square on the other leg. EXERCISES. 195 THEOEEMS. Ex. 356. The square constructed upon the sum of two straight lines is equivalent to the sum of the squares constructed upon these two lines, increased by twice the recteingle of these lines. Let AB and BG be the two straight lines, and AC their sum. Con- -struct the squares ACGK and ABED upon AC and AB, respectively. Prolong BE and DE until they meet KG and GG, respectively. Then we have the square EFGH, with sides each equal to BG. Hence, the square ACGK is the sum of the squares ABED and EFGH, and the rectangles BEHK and BGFE, tlie dimensions of which are equal to AB and BG. B C K E G Ex. 357. The square constructed upon the difference of two straight lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines. Let AB and AG he the two straight lines, and BC their difference. Construct the square ABFG upon AB, the square jj jf ACKH upon AG, and the square BE DC upon BC (as shown in the figure). Prolong ED to meet AG'mL. The dimensions of the rectangles LEFG and HKDL are AB and AG, and the square BCDE is evidently the difference between the whole figure and the smn of these rectangles ; that is, the square constructed upon BC is equivalent to the sum of the squares constructed Q "' upon AB and jiC, diminished by twice the rectangle of AB and AC. Ex. 358. The difference between the squares constructed upon two straight lines is equivalent to the rectangle of the sum and difference of these lines. Let ABDE and BGGF be the squares constructed upon the two straight lines AB and BC. The difference between these squares is the .polygon AGGFDE, which is com- posed of the rectangles ACHE and GFDH. Prolong AE and GH to I and K, respectively, making EI and HK each equal to BC, and draw IK. The rectangles GFDH and EHKI are equal. The difference between the squares ABDE and BGGF is then equivalent to the rectangle ACKI, which has for dimensions AI, equal to AB + BC, and EH, equal to AB — BC. I JR H E G D V c s 196 BOOK IV. PLAira; geometry. Ex. 359. The area of a rhombus is equal to half the product of its diagonals. Ex. 360. Two isosceles triangles are equivalent If their legs are equal each to each, and the altitude of one is equal to half the base of the other. Ex. 361. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. Ex. 362. Two parallelograms are equal if two adjacent sides of the one are equal, respectively, to two adjacent sides of the other, and the included angles are supplementary. Ex. 363. If ABC is a right triangle, C th e verte x of the right angle, BD a line cutting AC hi D, then BD^ + AC'^ = AB^ + DC'^. Ex. 364. Upon the sides of a right triangle as homologous sides three similar polygons are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the legs. Ex. 365. If the middle points of two adjacent sides of a parallelogram are joined, a triangle is formed which is equivalent to one eighth of tjie parallelogram. Ex. 366. If any point within a parallelogram is joined to the four ver- tices, the sum of either pair of triangles having parallel bases is equivalent to half the parallelogram. Ex. 367. Every straight line drawn through the intersection of the diagonals of a parallelogram divides the parallelogram into two equal parts. Ex. 368. The line which joins the middle points of the bases of a trape- zoid divides the trapezoid into two equivalent parts. Ex. 369. Every straight line drawn through the middle point of the median of a trapezoid cutting both bases divides the trapezoid into two equivalent parts: Ex. 370. If two straight lines are drawn from the middle point of either leg of a trapezoid to the opposite vertices, the triangle thus formed is equivalent to half the trapezoid. Ex. 371. The area of a trapezoid is equal to the product of one of the legs by the distance from this leg to the middle point of the other leg. Ex. 372. The figure whose vertices are the middle points of the sides of any quadrilateral is equivalent to half the quadrilateral. PROBLEMS OF CONSTRUCTION. 197 PROBLEMS OF CONSTRUCTION. Proposition XI. Problem. 417. To construct a square equivalent to the sum of two given squares. —\ -St.. R B' Let R and R' be two given squares. To construct a square equivalent to W + R. Construct the rt. Z A. Take AC equal to a side of W, and AB equal to a side of R ; and draw BC. Construct the square S, having each of its sides equal to BC. Then S is the square required. Proof. 'BC'' =0= AG'' + Zb", § 415 (iAe square on the hypotenuse of a rt. A is equivalent to the sum of the squares on the two legs). .-.S <^B' + B. Q. E. F. Ex. 373. If the perimeter of a rectangle is 72 feet, and the length is equal to twice the width, find the area. Ex. 374. How many tiles 9 inches long and 4 inches wide will be required to pave a path 8 feet wide surrounding a rectangular court 120 feet long and 36 feet wide ? Ex. 375. The bases of a trapezoid are 16 feet and 10 feet ; each leg is equal to 5 feet. Find the area of the trapezoid. 198 BOOK IV. PLANE GEOMETKY. Proposition XII. Problem. 418. To construct a square equivalent to the difference of two given squares. BK Let R be the smaller square and R' the larger. To construct a square equivalent to B' — R. Construct the rt. Z A. Take AB equal to a side of R. Prom 5 as a centre, 'with a radius equal to a side of R', describe an arc cutting the line AX at C. Construct the square S, having each of its sides equal to AC. Then S is the square required. Proof. AC'^BC AB. §416 (the square on either leg of a rt. A is equivalent to the difference of the square on the hypotenuse and the square on the other leg). .*. O =0= Jh si. Q. E. F. Ex. 376. Construct a square equivalent to the sum of two squares ■whose sides are 3 inches and 4 iuclaes. Ex. 377. Construct a square equivalent to the difference of two squares whose sides are 2J inches and 2 inches. Ex. 378. Find the side of a square equivalent to the sum of two squares whose sides are 24 feet and 32 feet. Ex. 379. Find the side of a square equivalent to the difference of two squares whose sides are 24 feet and 40 feet. Ex. 380. A rhomhus contains 100 square feet, and the length of one diagonal is 10 feet. Find the length of the other diagonal. ' PROBLEMS OP CONSTRXJCTION. 199 Peoposition XIII. Pboblem". 419. To construct a polygon similar to two given simi- lar polygons and equivalent to their sum. n" I 0^. "N, A"' B" P H Let R and R' be two similar polygons, and AB and A'B' two homologous sides. To construct a stTnilar polygon equivalent to It + R'. Construct the rt. Z. P. Take PH equal to A'B', and PO equal to AB. Draw OH, and take A"B" equal to OH. Upon A"B", homologous to AB, construct B" similar to B. Then B" is the polygon required. Proof. PO' + PH^ = '0H\ % 415 Put for PO, PH, and OH their equals AB, A'B', and A"B"^ Then Now — = '"~" „ , and ^^ = ^^ „ • § 412 AB' + A'B''' = A"B"\ B^ B" A^ A"B"'' and B' A'B'" B" A"B"'' B + B' AB' + A'B'" _ ^ By addition, -^^ — = — „ — = 1. Ax. 2 B" A"B"^ .-.B" ^B + B'. Q.E.F, 200 BOOK IV. PLANE GEOMETRY. Proposition XIV. Problem. 420. To construct a triangle equivalent to a given polygon. c D 1 A E F ^^ Let ABCDHE be the given polygon. To construct a triangle equivalent to the given polygon. Let D, H, and E be any three consecutive vertices of the polygon. Draw the diagonal DE. Prom H draw HF II to DE. Produce AE to meet HF at F, and draw DF. Again, draw CF, and draw DK II to CF to meet AF pro- duced at K, and draw CK. In like manner continue to reduce the number of sides of the polygon until we obtain the A CIK. Then A CIK is the triangle required. Proof. The polygon ABCDF has one side less than the poly- gon ABCDHE, but the two polygons are equivalent. Por the part ABCDE is common, and the A DEE =o A DEH, § 404 (for the base DE is common, and their vertices F and H are in the line FS II to the base). In like manner it may be proved that ABCE =0= ABCDF, and CIK =o= ABCK. q.b.f. PROBLEMS OF CONSTRUCTION. 201 Proposition XV. Problem. 421. To construct a square equivalent to a given par- allelogram. 1 J jJu N Let ABCD be the parallelogram, b its base, and a its altitude. To construct a square equivalent to the O ABCD. Upon a line MX take MN equal to a, NO equal to b. Upon MO as a diameter, describe a semicircle. At N erect NP _L to MO, meeting the circumference at P. Then the square R, constructed upon a line equal to NP, is equivalent to the O ABCD. Proof. MN ■.NP = NP: NO, § 370 (a ± let fall from any point of a circumference to the diameter is the mean proportional between the segments of the diameter). .-. NP^ = MN xNO = aXb. § 327 Therefore, B ^ EJ ABCD. q.e.f. 423. Cor. 1. A square may he constructed equivalent to a given triangle, hy taking for its side the mean proportional between the base and half the altitude of the triangle. 423. Cor. 2. A square may be constructed equivalent to a given polygon, by first reducing the polygon to an equivalent triangle, and then constructing a square equivalent to the triangle. 202 BOOK IV. PLANE GEOMETRY. Proposition XVI. Problem. 424. To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line. R Let E be the given square, and let the sum of the base and alti- tude of the required parallelogram be equal to the given line MN. To construct a O equivalent to R, with the sum of its base and altitude equal to MN. Upon MN as a diameter, describe a semicircle. At M erect MP, a _L to MN, equal to a side of the given square R. Draw PQ II to MN, cutting the circumference at S. Draw SC ± to MN. Any O having CM for its altitude and CN for its base is equivalent to R. Proof. SC- 12 PM. §§ 104, 180 .■.SC' = PM^ = R. But MC:SC= SC: CN, § 370 (a J_ let fall from any point of a circumference to the diameter is the mean proportional between the segments of the diameter). Then 'sc' :o= 3IC X CN. § 327 Q.E.F. Note. This problem may be stated as follows : To construct two straight lines the sum and product of which are knowru PROBLEMS OF CONSTIlUCTIOlSr. 203 Proposition XVII. Pkoblem. 425. To 'construct a parallelogram equivalent to a given square, and having the difference of its hose and altitude equal to a given line. y Let R be the given square, and let the difference of the base and altitude of the required parallelogram be equal to the given line MN. To construct a O equivalent to It, with the difference of the base and altitude equal to MN. Upon the given line MN as a diameter, describe a circle. From M draw MS, tangent to the O, and equal to a side of the given square R. Through the centre of the O dra-w SB intersecting the cir- cumference at C and B. Then any O, as R', having SB for its base and SO for its altitude, is equivalent to B. Proof. SB:SM=SM: SC, § 381 (if from a point without aQ a secant and a tangent are drawn, th£ tangent is the mean proportional between the whole secant and the external segment). Then SM"" ^ SB X SO, § 327 and the difference between SB and SC is the diameter of the O, that is, MN. q.e.p. Note. This problem may he stated : To construct two straight lines the difference and product of which are known. 204 BOOK IV. PLANE GEOMETRY. Proposition XVIII. Problem. 426. To construct a polygon similar to a given poly- gon P and equivalent to a given polygon Q. p' / Let P and Q be the two given polygons, and AB a side of P. To construct a polygon similar to P and equivalent to Q. Find squares equivalent to P and Q, § 423 and let m and n respectively denote their sides. Find A'B', the fourth proportional to m, n, and AB. § 386 Upon A'B', homologous to AB, Construct P' similar to P. Then Proof. P'-Q. m:n = AB: A'B'. Const. .■.m^:n'' = AB":A'B'\ § 338 P =c= m% and Q =c= ral Const. P:Q = m^:n^ = AB" : A'B'". P-.P'^AB': A'B'\ §412 .•.P:Q = P:P'. Ax. 1 .: P' ^ Q. Q.E.F, But But Ex. 381. To construct a square equivalent to the sum of any number of given squares. Ex. 382. To construct a polygon similar to two given similar polygons and equivalent to their difference. PBOBLEMS OF CONSTRUCTION. 205 Pboposition XIX. Pkoblbm. 427. To construct a square which shall have a given ratio to a given square. L-it- -)0 "^ Let R be the given square, and — the given ratio. To construct a square which shall be to S as n is to rti. Take AB equal to a side of B,, and draw Ay, making any convenient angle with AB. On Ay take AB equal to m, EF equal to n, and draw BB. Draw FC II to EB meeting AB produced at C. On j4C as a diameter, describe a semicircle. At B erect the _L BD, meeting the semicircumf erence at D. Then BD is a side of the square required. Proof. Denote AB \)j a, BC by b, and BD by x. Now a:x = x:b. § 370 Therefore, a^ : x^ = a : b. § 337 But a:b = m:n. § 342 Therefore, a" : x^ = m : n. Ax. 1 By inversion, x^ : a^ '= n : m. § 331 Hence, the square on BD wUl have the same ratio to R as w has to w. Q.E.F. 206 BOOK IV. PLANE GEOMETRY. Proposition XX. Problem. 428. To construct a polygon similar to a given poly- gon and having a given ratio to it. \ \ \ \ J ^ ;g, Let R be the given polygon, and — the given ratio. To construct a 2>olygo7i similar to R, which shall be to R as n is to m. Construct a line A'B', such that the square on A'B' shall be to the square on AB as n is to m. § 427 Upon A'B', as a side homologous to AB, construct the poly- gon S similar to B. § 391 Then S is the polygon required. Proof. S:R = A'B'-: ABT. § 412 But AJB''' : AJ? = n : m. Const. Therefore, S : R = n : m. Ax. 1 Q.E.F. Ex. 383. To construct a triangle equivalent to a given triangle, and having one side equal to a given length I. Ex. 384. To transform a triangle into an equivalent right triangle. Ex. 385. To transform a given triangle into an equivalent right tri- angle, having one leg equal to a given length. Ex. 386. To transform a given triangle into an equivalent right tri- angle, having the hypotenuse equal to a given length. EXERCISES. 207 PROBLEMS OF CONSTEUCTION. Ex. 387. To transform a triangle ABC into an equivalent triangle, having a side equal to a given length I, and an angle equal to angle BAC. Upon AB (produced if necessary), take AB equal to I, draw BE II to CD, meeting AC (produced if necessary) at E. A BED •^ A BEC. Ex. 388. To transform a given triangle into an equivalent isosceles tri- angle, having the base equal to a given length. To construct a triangle equivalent to : Ex. 389. The sum of two given triangles. Ex. 390. The difference of two given triangles. Ex. 391. To transform a given triangle into an equivalent equilateral triangle. To transform a parallelogram into an equivalent : Ex. 392. Parallelogi-am having one side equal to a given length. Ex. 393. Parallelogram having one angle equal to a given angle. Ex. 394. Rectangle having a given altitude. To transform a square into an equivalent : Ex. 395. Equilateral triangle. Ex. 396. Right triangle having one leg equal to a given length. Ex. 397. Rectangle having one side equal to a given length. To construct a square equivalent to : Ex. 398. Five eighths of a given square. Ex. 399. Three fifths of a given pentagon. Ex. 400. To divide a given triangle iuto two equivalent parts by a line through a given poiat P in one of the sides. Ex. 401. To find a point within a triangle, such that the lines joining this point to the vertices shall divide the triangle into three equivalent parts. Ex. 402. To divide a given triangle into two equivalent parts by a line parallel to one of the sides. Ex. 403. To divide a given triangle into two equivalent parts by a line perpendicular to one of the sides. y08 BOOK IV. PLANE GEOMETRY. PROBLEMS OF COMPUTATION. Ex. 404. To find the area of an equilateral triangle in terms of its side. Denote the side hy a, the altitude hy h, and the area by S. Then /j2^a2_^=l^ = « x3. §372 4 4 4 But .■.h = ^V3. 2 2 a aVs _ ^2 ^ 2 " 4 §403 Ex. 405. To find the area of a triangle in terms of its sides. By Ex. 312, /j = - Vs(s - a) (s - 6) (s - c). Vs (8 -a)(s- 6) (s - c). Ex. 406. To find the area of a triangle in terms of the radius of the circumscribed circle. If E denotes the radius of the circumscribed circle, and h the altitude of the triangle, we have, by § 384, 6xc = 2BxA. Multiply by a, and we have, ax6xc = 2ExaxA. But ax h = 2S. .-. a X b X c = 4 B X S. abc ^■" "~--.v A / a-- '- / V\ /' ^^ § 403 \ ^^r ,y' ..s = - in Show that the radius of the circumscribed circle is equal to abe 4S' EXERCISES. 209 Ex. 407. Find the area of a right triangle, if the length of the hypote- nuse is 17 feet and the length of one leg is 8 feet. Ex. 408. Eind the ratio of the altitudes of two equivalent triangles, if the base of one is three times that of the other. Ex. 409. The bases of a trapezoid are 8 feet and 10 feet, and the alti- tude is 6 feet. Eind the base of the equivalent rectangle that has an equal altitude. Ex. 410. Eind the area of a rhombus, if the sum of its diagonals is 12 feet, and their ratio is 3 : 5. Ex. 411. Eind the area of an isosceles right triangle, if the hypotenuse is 20 feet. Ex. 412. In a right triangle the hypotenuse is 13 feet, one leg is 5 feet. Eind the area. Ex. 413. Find the area of an isosceles triangle, if base = 6, and leg = c. Ex. 414. Find the area of an equilateral triangle, if one side = 8 feet. Ex. 415. Eind the area of an equilateral triangle, if the altitude = h. Ex. 416. A house is 40 feet long, 30 feet wide, 25 feet high to the roof, and 35 feet high to the ridge-pole. Find the number of square feet in its entire exterior surface. Ex. 417. The sides of a right triangle are as 3 : 4 : 5. The altitude upon the hypotenuse is 12 feet. Eind the area. Ex. 418. Find the area of a right triangle, if one leg = o, and the alti- tude upon the hypotenuse = h. Ex. 419. Find the area of a triangle, if the lengths of the sides are 104 feet, 111 feet, and 175 feet. Ex. 420. The area of a trapezoid is 700 square feet. The bases are 30 feet and 40 feet, respectively. Find the altitude. Ex. 421. ABGB is a trapezium ; AB = ?n feet, BG = 119 feet, CD = 41 feet, DA = 169 feet, ^O = 200 feet. Eind the area.' Ex. 422. What is the area of a quadrilateral circumscribed about a circle whose radius is 25 feet, if the perimeter of the quadrilateral is 400 feet ? What is the area of a hexagon that has a perimeter of 400 feet and is circumscribed about the same circle of 25 feet radius (Ex. 361) ? Ex. 423. The base of a triangle is 15 feet, and its altitude is 8 feet. Find the perimeter of an equivalent rhombus, if the altitude is 6 feet. 210 BOOK IV. PLANE GEOMETRY. Ex. 424. Upon the diagonal of a rectangle 24 feet by 10 feet a triangle equivalent to the rectangle is constructed. What is its altitude ? Ex. 425. Find the side of a squai-e equivalent to a trapezoid whose bases are 56 feet and 44 feet, and each leg is 10 feet. Ex. 426. Through a point P ia the side AB of a triangle ABC, a line is drawn parallel to BC so as to divide the triangle into two equivalent parts. Fiud the value of AF in terms of AB. Ex. 427. What part of a parallelogram is the triangle cut off by a liue from one vertex to the middle point of one of the opposite sides ? Ex. 428. In two similar polygons, two homologous sides are 15 feet and 25 feet. The area of the first polygon is 450 square feet. Find the area of the second polygon. Ex. 429. The base of a triangle is 32 feet, its altitude 20 feet. What is the area of the triangle out off by a line parallel to the base at a distance of 15 feet from the base ? Ex. 430. The sides of two equilateral triangles are 3 feet and 4 feet. Fiud the side of an equilateral triangle equivalent to their sum. Ex. 431. If the side of one equilateral triangle is equal to the altitude of another, what is the ratio of their ai-eas ? Ex. 432. The sides of a triangle are 10 feet, 17 feet, and 21 feet. Fiud the areas of the parts into which the triangle is divided by the bisector of the angle formed by the first two sides. Ex. 433. In a trapezoid, one base is 10 feet, the altitude is 4 feet, the area is 32 square feet. Find the length of a line drawn between the legs parallel to the bases and distant 1 foot from the lower base. Ex. 434. The diagonals of a rhombus are 90 yards and 120 yards, respectively. Find the area, the length of one side, and the perpendicu- lar distance between two parallel sides. Ex. 435. Find the number of square feet of carpet that ai-e required to cover a triangular floor whose sides are, respectively, 26 feet, 35 feet, and 51 feet. Ex. 436. If the altitude h of & triangle is increased by a length m, how much must be taken from the base a that the area may remain the same ? Ex. 437. Find the area of a right triangle, having given the segments p, q, into which the hypotenuse is divided by a perpendicular drawn to the hypotenuse from the vertex of the right angle. Book V. eegulae, polygolfs and ciecles. 429. Def. a regular polygon is a polygon which is both equilateral and equiangular. The equilateral triangle and the square are examples. Proposition I. Theoeem. 430. An equilateral iMlygon inscribed in a circle is a regular polygon. Let ABC etc. be an equilateral polygon inscribed in a circle. To prove that the polygon ABC etc. is a regular polygon. Proof. The arcs AB, BC, CD, etc., are equal. § 243 Hence, arcs ABC, BCD, etc., are equal. Ax. 2 Therefore, arcs CFA, DFB, etc., are equal. Ax. 3 Therefore, A A, B, C, etc., are equal. § 289 Therefore, the polygon ABC etc. is a regular polygon, being equilateral and equiangular. § 429 Q.E.D. 211 212 BOOK V. PLANE GEOMETRY. Pkoposition II. Theokem. 431. A circle may he circumscribed about, and a circle m,ay be inscribed in, any regular polygon. Let ABCDE be a regular polygon. 1. To prove that a circle may he circumscribed about ABCDE. Proof. Let be the centre of the circle which may be passed through A, B, and G. Draw OA, OB, OC, and OD. Then ZABC = ZBCI), and By subtraction, /LOBC = AOCB. 258 §429 §145 Ax. 3 Z OBA = Z OCD. The A OBA and OCD are equal § 143 For Z OBA = Z.OCD, OB = OC, § 217 and AB = CD. § 429 .;. OA = OD. § 128 .'. the circle passing through A, B, C, passes through D. In like manner it may be proved that the circle passing through B, C, and D also passes through E; and so on. REGULAR POLYGONS AND CIRCLES. 213 Therefore, the circle described from as a centre, with a radius OA, will be circumscribed about the polygon. § 231 2. To prove that a circle -may he inscribed in ABODE. Proof. Since the sides of the regular polygon are equal chords of the cii'ciiniscribed circle, they are equally distant from the centre. § 249 Therefore, the circle described from as a centre, with the distance from to a side of the polygon as a radius, will be inscribed in the polygon (§ 232). q.e.d. 432. Def. The radius of the circumscribed circle, OA, is called the radius of the polygon. 433. Dee. The radius of the inscribed circle, OF, is called the apothem of the polygon. 434. Dee. The common centre, 0, of the circumscribed and inscribed cii'cles is called the centre of the polygon. 435. Dep. The angle between radii drawn to the extremi- ties of any side is called the angle at the centre of the polygon. By joining the centre to the vertices of a regular polygon, the polygon can be decomposed iato as many equal isosceles triangles as it has sides. 436. CoK. 1. TJie angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon. Hence, the angles at the centre of any regu- lar polygon are all equal. 437. CoE. 2. The radius drawn to any vertex of a regular polygon bisects the angle at the vertex. 438. CoK. 3. The angle at the centre of a regular polygon and an interior angle of the polygon are supplementary. For A FOB and FBO are complementary. § 135 .'. their doubles AOB and FBO are supplementary. Ax. 6 214 BOOK V. PLANE GEOMETRY. Pkoposition III. Theorem. 439. If the circumference of a circle is divided into any number of equal arcs, the chords joining the suc- cessive points of division form a regular inscribed poly- gon ; and the tangents drawn at the points of division form a regular circumscribed piolygon. Suppose the circumference divided into equal arcs AB, BC, etc. Let AB, BC, etc., be the chords, FBG, GCH, etc., the tangents. 1. To prove that ABODE is a regular polygon. Proof. The sides AB, BC, CD, etc., are equal. § 241 Therefore, the polygon is regular. § 430 2. To prove that FGHIK is a regular polygon. Proof. The A AFB, BGC, CHD, etc., ai-e all equal isosceles triangles. §§ 295, 139 .". A F, G, H, etc., are equal, and FB, BG, GC, etc., are equal. .-. FG = GH= HI, etc. Ax. 6 .-. FGHIK is a regulai polygon. § 429 Q.E.D. 440. CoR. 1. Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides as the inscribed polygon. REGULAR POLYGONS AND CIRCLES. 215 441. CoK. 2. Tangents to a circle at the middle points of the arcs subtended hy the sides of a regular inscribed polygon form a circumscribed regular polygon, whose sides are parallel to the sides of the inscribed polygon and ivhose vertices lie on the radii (^prolonged) of the in^ scribed polygon- For two corresponding sides, AB and A'B', are perpendicular to OM (§§ 248, 254), and are parallel (§ 104) ; and the tangents MB' and NB', intersecting at a point equidistant from OM and ON (§ 261), intersect upon the bisector of the Z MON (§ 162) ; that is, upon the radius OB. § 4~37 442. CoE. 3. If the vertices of a regular inscribed polygon are joined to the middle points of the arcs sub- tended hy the sides of the polygon, the joining lines form a regular inscribed polygon of kI double the number of sides. 443. CoE. 4. Tangents at the middle points of the arcs between adjacent points of contact of the sides of a regular circumscribed poly- gon form a regular circumscribed polygon of m doubleithe number of sides. 444. Cor. 5. The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double the number of sides; and the perimeter of a circumscribed polygon is greater than the perimeter of a circumscribed polygon of double the number of sides. For two sides of a triangle are together greater than the third side. § 138 216 BOOK V. PLANE GEOMETRY. Proposition IV. Theorem. 445. Two regular polygons of the same number of sides are similar. P' D JE'^ No' A B A' B Let Q and Q' be two regular polygons, each having n sides. To prove that Q and Q' are similar. Proof. The sum of the interior A of each polygon is equal to (n- 2)2 It. A, §205 (the sum of the interior A of a polygon is equal to 2 rt. A taken as many times less two as the polygon has sides). (n — 2)2 rt. A Each angle of either polygon = -^^ > § 206 (for the A of a regular polygon are all equal, and hence each Z is equal to the sum of the A divided by their number). Hence, the two polygons Q and Q' axe mutually equiangular. Since AB = BC, etc., and A'B' = B'C, etc., § 429 AB:A'B' = BC:B'C',etG. Hence, the two polygons have their homologous sides pro- portional. Therefore the two polygons are similar. § 351 Q.E.D. 446. Cor. The areas of two regular polygons of the same number of sides are to each other as the squares of any two homologous sides. § 412 REGUXAR POLYGONS AND CIRCLES. 217 Proposition V. Theorem. 447. The perimeters of two regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, and also as the radii of their inscribed circles. M B M' B' Let P and P' denote the perimeters, and 0' the centres, of the two regular polygons. From 0, 0' draw OA, O'A', OB, O'B', and the Js OM, O'M'. To prove that P:P< = 0A: O'A' = OM : O'M'. Proof. Since the polygons are similar, §445 P:P' = AB: A'B'. §364 The A GAB and O'A'B' are isosceles. §431 Now Z.O = Z.O', §436 id OA:OB = 0'A':0'B'. .-. the A OAB and O'A'B' are similar. §357 .■.AB:A'B' = OA:0'A'. §351 Also, AB : A'B' = OM : O'M'. §361 .-. P:P' = OA: O'A' = OM : O'M'. Ax. 1 Q.E.D. 448. Cor. The areas of two regular polygons of the same number of sides are to each other as the squares of the radii of the circumscribed circles, and of the inscribed circles. § 413 218 BOOK V. PLANE GEOMETRY. Proposition VI. Theorem. 449. If the number of sides of a regular inscribed polygon is indefinitely increased, the apothem of the polygon approaches the radius of the circle as its limit. Let AB be a side and OP the apothem of a regular polygon of n sides inscribed in the circle whose radius is OA. To prove that OP approaches OA as a limit, when n in- creases indefinitely. Proof. OP < OA, § 97 and OA-OPX always, imi SJ R^ - i a?. The Z DAH is a rt. Z. § 290 In the rt. A DAH, AD = DH Y. DC. § 367 But DH=2R, and DC = R - ^R" ~ ia''. .-. AD = V2i?(i?- Vi?2-^a=f) 479. Cob. If R = 1, AD = '\/2 - ^4:- a". Q. E. F, MAXIMA AND MINIMA. 235 Proposition XXI. Theorem. 485. Of all isoperimetric triangles having the same hose the isosceles triangle is the Tnaximum. Let the A ACB and ADB have equal perimeters, and let AC and CB be equal, and AD and DB be unequal. To prove that A ACB > A ADB. Proof. Produce AC to H, making CII= AC; and draw HB. Produce HB, take DP equal to DB, and draw AP. Draw CE and DF J_ to AB, and CK and DM II to AB. The Z. ABH is a right Z., for it may be inscribed in the semicircle whose centre is C and radius CA. § 290 ADP is not a straight line, for then the A DBA and DAB would be equal, being complements of the equal A DBM and DPM, respectively ; and DA and DB would be equal (§ 147), which is contrary to the hypothesis. Hence, APKAD + DP, .-. BP. §102 .-. CE(= i: BH) > DF(= i BP). Ax. 7 Therefore, A ACB > A ADB. § 405 Q.E.D. 236 BOOK V. PLANE GEOMETRY. Proposition XXII. Theoeem. 486. Of all polygons with sides all given hut one, the maximum can he inscribed in a semicircle which has the undetermined side for its diameter. Let ABCDE be the maximum of polygons with sides AB, BC, CD, DE, and the extremities A and E on the straight line MN. To prove that ABCDE can be inscribed in a semicircle. Proof. From any vertex, as C, draw CA and CE. The A ACE must be the maximum of all A having the sides CA and CE, and the third side on MN; otherwise, by increasing or diminishing the Z. A CE, keeping the lengths of the sides CA and CE unchanged, but sliding the extremities A and E along the line MN, we coiild increase the A ACE, while the rest of the polygon would remain unchanged ; and therefore increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence, the A A CE is the maximum of A that have the sides CA and CE. Therefore, the Z ACE is a. right angle. § 484 Therefore, C lies on the semicircumference. § 290 Hence, every vertex lies on the circumference ; that is, the maximum polygon can be inscribed in a semicircle having the undetermined side for a diameter. o e d MAXIMA AND MINIMA. 237 Proposition XXIII. Theorem. 487. Of all polygons loith given sides, that which can be inscribed in a circle is the maximum. let ABCDE be a polygon inscribed in a circle, and A'B'C'D'E' be a polygon, equilateral with respect to ABCDE, which cannot be in- scribed in a circle. To prove that ABCDE > A'B'C'D'E'. Proof. Draw the diameter AH, and draw CH and DH. Upon CD construct the A C'H'D' = A CHD, and draw^'^'. Since, by hypothesis, a O cannot pass through all the vertices of A'B'C'H'D'E', one or both of the parts ABCH, AEDH must be greater than tlie corresponding part of A'B'C'H'D'E'. § 486 If either of these parts is not greater than its corresponding part, it is equal to it, § 486 (for ABCH and AEDH are the maxima of polygons that have sides equal to AB, BG, CH, and AE, ED, DH, respectively, and the remaining side undetermined). .-.AB CHDE > A'B' C'H'D'E'. Ax. 4 Take away from the two figures the equal A CHD and CHD'. Then ABCDE > A'B'C'D'E'. Ax. 5 Q.E.D. 236 BOOK V. PLANE GEOMETRY. Peoposition XXII. Theokem. 486. Of all polygons with sides all given hut one, the maximum can be inscribed in a semicircle which has the undetermined side for its diameter. Let ABCDE be the maximum of polygons with sides AB, BC, CD, DE, and the extremities A and E on the straight line MN. To prove that ABODE can be inscribed in a semicircle. Proof. Erom any vertex, as C, draw CA and CH. The A ACH must be the maximum of aJl A having the sides CA and CE, and the third side on MN; otherwise, by increasing or diminishing the Z. A CJE, keeping the lengths of the sides CA and CE unchanged, but sliding the extremities A and E along the line JfJV, we could increase the A ACE, while the rest of the polygon would remain unchanged ; and therefore increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence, the A A CE is the maximum of A that have the sides CA and CE. Therefore, the Z A CE is a right angle. § 484 Therefore, C lies on the semicircumference. § 290 Hence, everT/ vertex lies on the circumference ; that is, the maximum polygon can be inscribed in a semicircle having the undetermined side for a diameter. r, ^ ^ y. ti, Ut MAXIMA AND MINIMA. 237 Proposition XXllI. Theorem. 487. Of all polygons loith given sides, that which can he inscribed in a circle is the maximum. Let ABCDE be a polygon inscribed in a circle, and A'B'C'D'E' be a polygon, equilateral with respect to ABCDE, which cannot be in- scribed in a circle. To prove that AB CDE > A'B'C'D'E'. Proof. Draw the diameter AH, and draw CH and DH. Upon CD' construct the A CH'!)' = A CUD, and draw^'^'. Since, by hypothesis, a O cannot pass through all the vertices of A' B' C H' D' E' , one or both of the pai-ts ABCH, AEDH must be greater than tlie corresponding part of A'B'C'H'D'E'. § 486 If either of these parts is not greater than its corresponding part, it is equal to it, § 486 (for ABCH and AEDH are the maxvna of polygons that have sides equal to AB, BG, CH, and AH, ED, DH, respectively, and the remaining side undetermined). .-. ABCHDE > A'B'C'H'D'E'. Ax. 4 Take away from the two figures the equal A CHD and C'HD'. Then ABCDE > A'B'C'D'E'. Ax. 5 Q.E.D. 238 BOOK V. PLANE GEOMETRY. Proposition XXIV. Theorem. 488. Of isoperimetrie polygons of the same number of sides, the maximum is equilateral. 5 K Let AECD etc. be the maximum of isoperimetrie polygons of any given number of sides. To 2)rov6 that ^AB, BC, CD, etc., are equal. Proof. Draw AG. The A ABC must be the maximum of all the A which ai'e formed upon AC with a perimeter equal to that of A ABC. Otherwise a greater A .4 JfC could be substituted for A ^i?C, without changing the perimeter of the polygon. But this is inconsistent with the hypothesis that the poly- gon ABCD etc. is the maximum polygon. .-. the A ABC is isosceles. § 485 .■.AB = BC. In like manner it may be proved that BC = CD, etc. q.e.d. 489. CoE. The maximum of isoperimetrie polygons of the same number of sides is a regular polygon. Por the maximum polygon is equilateral (§ 488), and can be inscribed in a circle (§ 487), and is, therefore, regular. § 430 Q. £. D. MAXIMA AND MINIMA. 239 Proposition XXV. Theorem. 490. Of isoperimetric regular polygons, that which has the greatest number of sides is the maximum. D B Let Q be a regular polygon of three sides, and Q' a regular poly- gon of four sides, and let the two polygons have equal perimeters. To prove that Q' is greater than Q. Proof. Draw CD from C to auy point in AB. Invert the A CDA and place it in tlie position DOE, letting I) fall at C, G at B, and A at E. The polygon DBCE is an irregular polygon of four sides, which by construction has the same perimeter as Q', and the same area as Q. Then the irregular polygon DBCE of four sides is less than the isoperimetric regular polygon Q' of four sides. § 489 In like manner it may be shown that Q' is less than an iso- perimetric regular polygon of five sides, and so on. q.e.d. Ex. 445. Of all equivalent parallelograms that have equal bases, the rectangle has the minimum perimeter. Ex. 446. Of all equivalent rectangles, the square has the minimum perimeter. Ex. 447. Of all triangles that have the same base and the same alti- tude, the isosceles has the .minimum perimeter. Ex. 448. Of all triangles that can be inscribed in a given circle, the equilateral is the ma.Trimnm and has the maximum perimeter. 240 BOOK V. PLANE GEOMETRY. Proposition XXVI. Theorem. 491. Of regular polygons having a given area, that which has the greatest number of sides has the least perimeter. Q' Let Q and Q' be regular polygons having the same area, and let Q' have the greater number of sides. To prove the perimeter ofQ> the perimeter of Q'. Proof. Let Q" be a regular polygon having the same perim- eter as Q', and the same number of sides as Q. Then Q' > Q", § 490 (of isoperimetric regular polygons, that which has the greatest number of sides is the maximum). But Q - Q'. .-. Q > Q". .'. the perimeter oi. Q > the perimeter of Q" But the perimeter of Q' = the perimeter of Q" .'. the perimeter oi Q > the perimeter of Q'. Hyp. Hyp. Q. E. D. Ex. 449. To inscribe in a semicircle the maximum rectangle. Ex. 450. Of all polygons of a given number of sides which may be inscribed in a given cii'cle, tliat ■which is regular has the maximum area and the maximum perimeter. Ex. 451. Of all polygons of a given numbeir of sides vhich may be circumscribed about a given circle, that which is regular has the TniniTniim area and the minimum perimeter. BXERCISES. 241 THBOBEMS. Ex. 452. Every equUateral polygon circumscribed about a circle is regula-r if it has an odd number of sides. Ex. 453. Every equiangular polygon inscribed in a. circle is regular i£ it has an odd number of sides. Ex. 454. Every equiangular polygon circumscribed about a circle is regular. Ex. 455. The side of a circumscribed equUateral triangle is equal to twice the side of the similar inscribed triangle. Ex. 456. The apothem of an inscribed regular hexagon is equal to half the side of the inscribed equilateral triangle. Ex. 457. The area of an iuscribed regular hexagon U three fourths of the area of the circumscribed regular hexagon. Ex. 458. The area of an inscribed regular hexagon is the mean propor- tional between the areas of the inscribed and the circumscribed equilateral triangles. Ex. 459. The square of the side of an inscribed equilateral triangle is equal to three times the square of a side of the inscribed regular hexagon. Ex. 460. The area of an inscribed equilateral triangle is equal to half the area of the inscribed regular hexagon. Ex. 461. The square of the side of an inscribed equilateral triangle is equal to the sum of the squares of the sides of the inscribed square and of the inscribed regular hexagon. Ex. 462. The square of the side of an inscribed regular pentagon is equal to the sum of the squares of the radius of the circle and the side of the inscribed regular decagon. If R denotes the radius of a circle, and a one side of an inscribed regulax polygon, show that : Ex. 463. In a regular pentagon, a = j E V 10 — 2 v5. Ex. 464. In a regular octagon, a = B. V2 — "^2. Ex. 465. In a regular dodecagon, a = B V2 — "vS. Ex. 466. If two diagonals of a regular pentagon intersect, the longer segment of each is equal to a side of the pentagon. 242 BOOK V. PLANE GEOMETRY. Ex. 467. The apothem of an inscritied regular pentagon is equal to half the sum of the radius of the circle and the side of the inscribed regular decagon. Ex. 468. The side of an inscribed regular pentagon is equal to the hypotenuse of the right triangle which has for legs the radius of the circle and the side of the inscribed regular decagon. Ex. 469. The radius of an- inscribed regular polygon is the mean pro- portional between its apothem and the radius of the similar circumscribed regular polygon. Ex. 470. If squares are constructed outwardly upon the six sides of a regular hexagon, the exterior vertices of these squares are the vertices of a regular dodecagon. Ex. 471. If the alternate vertices of a regular hexagon are joined by straight lines, show that another regular hexagon is thereby formed. Pind the ratio of the areas of these two hexagons. Ex. 472. If on the legs of a right triangle as diameters semicircles are described external to the triangle, and from the whole figure a semicircle on the hypotenuse is subtracted, the remaining figure is equivalent to the given right triangle. Ex. 473. The star-shaped polygon, formed by producing the sides of a regular hexagon, is equivalent to twice the given hexagon. Ex. 474. The sum of the perpendiculars drawn to the sides of a regular polygon from any point within the polygon is equal to the apothem multi- plied by the number of sides. Ex. 475. If two chords of a circle are perpendicular to each other, the sum of the four circles described on the four segments as diameters is equivalent to the given circle. Ex. 476. If the diameter of a circle is divided into any two segments, and upon these segments as diameters semicircumferences are described upon opposite sides of the diameter, these semicircumferences divide the circle into two parts which have the same ratio as the two segments of the diameter. Ex. 477. The diagonals that join any vertex of a regular polygon to all the vertices not adjacent divide the angle at that vertex into as many equal parts less two as the polygon has sides. EXERCISES. 243 PROBLEMS OP CONSTRUCTION. Ex. 478. To circumscribe au equilateral triaugle about a given circle. Ex. 479. To circumscribe a square about a given circle. Ex. 480. To circumscribe a regular hexagon about a given circle. Ex. 481. To circumscribe a regular octagon about a given circle. Ex. 482. To circumscribe a regular pentagon about a given circle. Ex. 483. To draw through a given point a line so as to divide a given circumference into two parts having the ratio 3 : 7. Ex. 484. To construct a circumference equal to the sum of two given circumferences. Ex. 485. To construct a circumference equal to the difference of two given circumferences. Ex. 486. To construct a circle equivalent to the sum of two given circles. Ex. 487. To construct a circle equivalent to the difference of two given circles. Ex. 488. To construct a circle equivalent to three times a given circle. Ex. 489. To construct a circle equivalent to three fourths of a given circle. Ex. 490. To construct a circle whose ratio to a given circle shall be equal to the given ratio m : n. Ex. 491. To divide a given circle hj a concentric circumference into two equivalent parts. Ex. 492. To divide a given circle by concentric circumferences into five equivalent parts. Ex. 493. To construct an angle of 18° ; of 36° ; of 9°. Ex. 494. To construct an angle of 12° ; of 24° ; of 6°. To construct with a side of a given length : Ex. 495. An equilateral triangle. Ex. 499. A regular pentagon. Ex. 496. A square. Ex. 500. A regular decagon. " Ex, 497. A regular hexagon. Ex. 501. A regular dodecagon. Ex. 498. A regular octagon. Ex. 502. A regular pentedecagon. 244 BOOK V. PLANE GEOMETRY. PROBLEMS OF COMPUTATION. Ex. 503. Find the area of a circle whose radius is 12 inches. Ex. 504. Find the circumference and the area of a circle whose diameter is 8 feet. Ex. 505. A regular pentagon is inscribed in a circle whose radius is B. If the length of a side is a, find the apothem. Ex. 506. A regular polygon is inscribed in a circle w hose radiu s is B. If the length of a side is a, show that the apothem is jVliJs _ a^. Ex. 507. Find the area of a regular decagon inscribed In a circle whose radius is 16 inches. Ex. 508. Find the side of a regular dodecagon inscribed in a circle whose radius is 20 inches. Ex. 509. Find the perimeter of a regular pentagon inscribed in a circle whose radius is 25 feet. Ex. 510. The length of each side of a park in the shape of a regular decagon is 100 yards. Find the area of the park. Ex. 511. Find the cost, at $2 per yard, of building a wall around a cemetery in the shape of a regular hexagon, that contains 16,627.84 square yards. Ex. 512. The side of an inscribed regular polygon of n sides is 16 feet. Find the side of an inscribed regular polygon of 2 n sides. Ex. 513. If the radius of a circle is B, and the side of an inscribed regular polygon is a, show that the side of the similar circumscribed regu- lar polygon is = • Ex. 514. What is the width of the circular ring between two concentric circumferences whose lengtlis are 650 feet and 425 feet ? Ex. 515. Find the angle subtended at the centre by an arc 5 feet 10 inches long, if the radius of the circle is 9 feet 4 iaches. Ex. 516. The chord of a segment is 10 feet, and the radius of the circle is 16 feet. Find the area of the segment. Ex. 517. Find the area of a sector, if the angle at the centre is 20°, and the radius of the circle is 20 inches. EXERCISES. 245 Ex. 518. The chord of half an arc is 12 feet, and the radius of the circle is 18 feet. Find the height of the segment subtended by the whole iU'O. Ex. 519. Find the side of a square which is equivalent to a circle whose diameter is 35 feet. Ex. 520. The diameter of a circle is 15 feet. Find the diameter of a circle twice as large. Three times as large. Ex. 521. Ftad the radii of the concentric circumferences that divide a circle 11 laches iu diameter into five equivalent parts. Ex. 522. The perimeter of a regular hexagon is 840 feet, and that of a regular octagon is the same. By how many square feet is the octagon larger than the hexagon ? Ex. 523. The diameter of a bicycle wheel is 28 inches. How many revolutions does the wheel make in going 10 miles ? Ex. 524. Find the diameter of a carriage wheel that makes 264 revolu- tions in going half a mile. Ex. 525. The sides of three regular octagons are 6 feet, 7 feet, 8 feet, respectively. Find the side of a regular octagon equivalent to the sum of the three given octagons. Ex. 526. A circular pond 100 yards in diameter is surrounded by a walk 10 feet wide. Find the area of the walk. Ex. 527. The span (chord) of a bridge in the form of a cu-cular arc is 120 feet, and the highest point of the ai-ch is 15 feet above the piers. Find the radius of the arc. Ex. 528. Three equal circles are described each tangent to the other two. If the common radius is B, find the area contained between the circles. Ex. 529. Given p, P, the perimeters of regular polygons of n sides inscribed in and circumscribed about a given circle. Find p', P', the perimeters of regular polygons of 2 n. sides inscribed in and circumscribed about the given circle. Ex. 530. Given the radius R, and the apothem r of an inscribed regular polygon of n sides. Find the radius R' and the apothem r' of an isoperi- metrical regular polygon of 2 m sides. 246 BOOK V. PLANE GEOMETRY. MISCELLANEOUS EXERCISES. THEOREMS. Ex. 531. If two adjacent angles of a quadrilateral are right angles, the bisectors of the other two angles are perpendicular. Ex. 532. If two opposite angles of a quadrilateral are right angles, the bisectors of the other two angles are parallel. Ex. 533. The two lines that join the middle points of the opposite sides of a quadrilateral bisect each other. Ex. 534. The line that joins the feet of the perpendiculars dropped from the extremities of the base of an isosceles triangle to the opposite sides is parallel to the base. Ex. 535. If AD bisects the angle A of a triangle ABC, and BD bisect? the exterior angle CBF, then angle ADB equals one half angle A CB. Ex. 536. The sum of the acute angles at the vertices of a, pentagram (five-pointed star) is equal to two right angles. Ex. 537. The altitudes AD, BE, CF of the triangle ABC bisect the angles of the triangle DEF. Circles with AB, BC, AC as diameters will pass through E and D, E and F, D and F, respectively. Ex. 538. The segments of any straight line intercepted between the cir- cumferences of two concentric circles are equal. Ex. 539. If a circle is circumscribed about any triangle, the feet of the perpendiculars dropped from any point in the circumference to the sides of the triangle lie in one straight line. Ex. 540. Two circles are tangent internally at P, and a chord ^B of the larger circle touches the smaller circle at C. Prove that PC bisects the angle APB. Ex. 541. The diagonals of a trapezoid divide each other into segments which are proportional. Ex. 542. If through a point P in the circumference of a circle two chords are drawn, the chords and the segments between P and a chord parallel to the tangent at P are reciprocally proportional. MISCELLANEOUS EXERCISES. 247 Ex. 543. The perpendiculars from two vertices of a triangle upon the opposite sides divide each other into segments reciprocally proportional. Ex. 544. The perpendicular from any point of a circumference upon a chord is the mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord. Ex. 545. In an isosceles right triangle either leg is the mean propor- tional between the hypotenuse and the perpendicular upon it from the vertex of the right angle. Ex. 546. If two circles intersect in the points A and B, and through A any secant CAD is drawn limited by the circumferences at C and D, the straight lines BC, BD are to each other as the diameters of the circles. Ex. 547. The area of a triangle is equal to half the product of its perim- eter by the radius of the inscribed circle. Ex. 548. The perimeter of a triangle is to one side as the perpendicular from the opposite vertex is to the radius of the inscribed circle. Ex. 549. If three straight lines AA', BB', CC, drawn from the vertices of a triangle ABC to the opposite sides, pass through a common point within the triangle, then OA^ OR PC _, AA' BB' "^ CG'~ Ex. 550. ABC is a triangle, M the middle point of AB, P any point in AB between A and M. If MD is drawn parallel to PC, meeting BC at D, the triangle BPD is equivalent to half the triangle ABC. Ex. 551. Two diagonals of a regular pentagon, not drawn from a com- mon vertex, divide each other in extreme and mean ratio. Ex. 552. If all the diagonals of a regular pentagon are drawn, another regular pentagon is thereby formed. Ex. 553. The area of an inscribed regular dodecagon is eqnal to three times the square of the radius. Ex. 554. The area of a square inscribed in a semicircle is equal to two fifths the area of the square inscribed in the circle. Ex. 555. The area of a circle is greater than the area of any polygon of equal perimeter. Ex. 556. The circumference of a circle is less than the perimeter of any polygon of equal area. 248 BOOK V. PLANE GEOMETRY. PROBLEMS OF LOCI. Ex. 557. Find the locus of the centre of the circle inscribed in a tri- angle that has a given base and a given angle at the vertex. Ex. 558. Find the locus of the intersection of the altitudes of a triangle that has a given base and a given angle at the vertex. Ex. 559. Find the locus of the extremity of a tangent to a given circle, if the length of the tangent is equal to a given line. Ex. 560. Find the locus of a point, tangents drawn from which to a given circle form a given angle. Ex. 561. Find the locus of the middle point of a line drawn from a given point to a given straight line. Ex. 562. Find the locus of the vertex of a triangle that has a given base and a given altitude. Ex. 563. Find the locus of a point the sum of whose distances from two given parallel lines is equal to a given length. Ex. 564. Find the locus of a point the difference of whose distances from two given parallel lines is equal to a given length. Ex. 565. Find the locus of a point the sum of whose distances from two given intersecting lines is equal to a given length. Ex. 566. Find the locus of a point the difference of whose distances from two given intersecting lines is equal to a given length. Ex. 567. Find the locus of a point whose distances from two given points are in the given ratio m : n. Ex. 568. Find the locus of a point whose distances from two given parallel lines are in the given ratio m : n. Ex. 569. Find the locus of a point whose distances from two given intersecting lines are in the given ratio m : n. Ex. 570. Find the locus of a point the sum of the squares of whose distances from two given points is constant. Ex. 571. Find the locus of a point the difference of the squares of whose distances from two given points is constant. Ex. 572. Find the locus of the vertex of a triangle that has a given base and the other two sides in the given ratio m : n. MISCELLANEOUS EXERCISES. 249 PROBLEMS OF CONSTRUCTION. Ex. 573. To divide a given trapezoid into two equivalent parts by a line parallel to the bases. Ex. 574. To divide a given trapezoid into two equivalent parts by a line through a given point in one of the bases. Ex. 575. To construct a regular pentagon, given one of the diagonals. Ex. 576. To divide a given straight line into two segments such that their product shall be the maximum. Ex. 577. To find a point in a semicircumference such that the sum of its distances from the extremities of the diameter shall be the maximum. Ex. 578. To draw a common secant to two given cii^les exterior to each other such that the intercepted chords shall have the given lengths a, b. Ex. 579. To draw through one of the points of intersection of two intersecting circles a common secant which shall have a given length. Ex. 580. To construct an isosceles triangle, given the altitude and one of the equal base angles. Ex. 581. To construct an equilateral triangle, given the altitude. Ex. 582. To construct a right triangle, given the radius of the inscribed circle and the difference of the acute angles. Ex. 583. To construct an equilateral triangle so that its vertices shall lie in three given parallel lines. Ex. 584. To draw a line from a given point to a given straight line which shall be to the perpendicular from the given point as m : n. Ex. 585. To find a point within a given triangle such that the perpen- diculars from the point to the three sides shall be as the numbers m, n, p. Ex. 586. To draw a straight line equidistant from three given points. Ex. 587. To draw a tangent to a given circle such that the segment intercepted between the point of contact and a given straight line shall have a given length. Ex. 588. To inscribe a straight line of a given length between two given circumferences and parallel to a given straight line. 250 BOOK V. PLANE GEOMETRY. Ex. 589. To draw through a given point a straight line so that its dis- tances from two other given points shall he in a given ratio. Ex. 590. To construct a square equivalent to the sum of a given tri- angle and a given parallelogram. Ex. 591. To construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon. Ex. 592. To construct a pentagon similar to a given pentagon and equivalent to a given trapezoid. Ex. 593. To find a point whose distances from three given straight lines shall be as the numbers ni, n, p. Ex. 594. Given an angle and two points P and P' between the sides of the angle. To find the shortest path from P to P' that shall touch both sides of the angle. Ex. 595. To construct a triangle, given its angles and its area. Ex. 596. To transform a given triangle into a triangle similar to another given triangle. Ex. 597. Given three points A, B, C. To find a fourth point P such that the areas of the triangles APB, APC, BPC shall be equal. Ex. 598. To construct a triangle, given its base, the ratio of the other sides, and the angle included by them. Ex. 599. To divide a given circle into n equivalent parts by concentric circumferences. Ex. 600. In a given equilateral triangle to inscribe three equal circles tangent to each other, each circle tangent to two sides of the triangle. Ex. 601. Given an angle and a point P between the sides of the angle. To draw through P a straight line that shall form with the sides of the angle a triangle with the perimeter equal to a given length a. Ex. 602. In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to two sides of the square. Ex. 603. In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to one side of the square. SOLID GEOMETRY. Book YI. lines and planes in space. DEFINITIONS. 492. Def. a plane is a surface such that a straight line joining any two points in it lies wholly in the surface. A plane is understood to be indefinite in extent ; but is usually represented by a parallelogram lying in the plane. 493. Def. A plane is said to be determined by given lines or points, if no other plane can contain the given lines or points without coinciding with that plane. 494. CoE. 1. One straight line does not determine a plane. For a plane can be made to turn about any straight line AB in it, and thus assume as many different positions as we please. 495. Cor. 2. A straight line and a point not in the line determine a plane. For, if a plane containing a straight line AB and any point C not in AB is made to revolve either way about AB, it will no longer contain the point C. 496. Cob. 3. Three points not in a straight line determine a plane. For by joining two of the points we have a straight line and a point without it, and these determine the plane. § 495 251 252 BOOK VI. SOLID GEOMETRY. 497. Cor. 4. Two intersecting lines determine a plane. For the plane containing one of these lines and any point of the other line not the point of intersection is determined. §495 Mr 498. Cob. 6. Two parallel lines determine a plane. For two parallel lines lie in a plane (§ 103), and a plane con- taining either parallel and a point in the other is determined. § 496 C ^ D A B 499. Def. "When we suppose a plane to be drawn through given points or lines, we are said to pass the plane through the given points or lines. 500. Dei-. When a straight line is drawn from a point to a plane, its intersection with the plane is called its foot. 501. Def. A straight line is perpendicular to a plane, if it is perpendicular to every straight line drawn through its foot in the plane ; and the plane is perpendicular to the line. 502. Def. A straight line and a plane are parallel if they cannot meet, however far both are produced. 503. Def. A straight line neither perpendiculai- nor par- allel to a plane is s.aid to be oblique to the plane. LINES AND PLANES. 253 504. Def. Two planes are parallel if they cannot meet, how- ever far they are produced. 505. Def. The intersection of two planes contains all the points common to the two planes. LINES AND PLANES. Proposition I. Theorem. 506. If two planes cut each other, their intersection is a straight line. Let MN and PQ be two planes which cut one another. To prove that their intersection is a straight line. Proof. Let A and B be two points common to the two planes. Draw a straight line through the points A and B. Then the straight line AB lies in both planes. § 492 No point not in the line AB can be in both planes ; for one plane, and only one, can contain a straight line and a point without the line. § 495 Therefore, the straight line through A and B contains all the points common to the two planes, and is consequently the intersection of the planes. § 505 Q. E. D. 254 BOOK VI. SOLID GEOMETRY. Proposition II. Theorem. 507. If a straight line is perpendicular to each of tivo other straight lines at their p>oint of intersection, it is per- pendicular to the plane of the two lines. Let AB be perpendicular to BC and BD at B. To prove that AB is A- to the jjlane MN of these lines. Proof. Through B draw in MN any other straight line BE, and draw CD cutting BC, BE, BD, at C, E, and D. Prolong AB to F, making BE equal to AB, and join A and F to each of the points C, E, and D. Then BC and BD are each ± to AE at its middle point. .■:AC = EC, and AD = ED. § 160 .■.AACD = AECD. §150 .■.ZACD = ZECD. §128 That is, ZACE = Z ECE. Hence, the A ACE and ECE are equal. § 143 Por AC = EC, CE = CE, and Z ACE = Z ECE. .-. AE = EE; and BE is _L to AF at B. § 161 .■. AB is J- to any and hence every line in TifiV" through B. .-. AB is ± to ilfiV". § 501 Q.E.D. LINES AND PLANES. 255 Proposition III. Theoebm. 508. All the perpendiculars that can be drawn to a straight line at a given point lie in a plane ivhich is perpendicular to the line at the given point. Let the plane MN be perpendicular to AB at B. To prove that BE, any _L to AB at B, lies in MN. Proof. Let the plane containing AB and BE intersect MN in the line BE' ; then AB is ± to BE. § 601 Since in the plane ABE only one _L can be drawn to AB at B (§ 83), BE and BE coincide, and BE lies in MN. Hence, every X to AB at B lies in the plane MN. q e d. 509. CoK. 1. At a given point in a straight line only one plane perpendicular to the line can he drawn, 510. CoE. 2. Through a given external point, one plane can be drawn perpendicular to a line, and only one. Let ^C be the line, and the point. Draw OC 1. to AC, and CD ± to ^C. Then CO and CD determine a plane j^^ through ± to ^C. [ \^ ,,,--"' 0\ Only one such plane can be drawn ; / <^ ~~-^ for only one _L can be drawn to AC from / ""-Z> the point 0. § 96 W 256 BOOK VI. SOLID GEOMETRY. PEOPOSiTioif IV. Theorem. 511. Through a given point there can he one perpen- dicular to a given plane, and only one. Fig. 1. Fig. 2. Case 1. When the given point is in the given plane. Let A be the given point in the plane MN (Fig. 1). To prove that there can be one perpendicular to the plane MN at A, and only one. Proof. Thiough A draw in MN any line BC, and pass through A a plane ^ AD. Proof. The rt. AA13C and ABD are equal. § 144 For AB is common, and BC = BD. Hyp. .■.AC = AD. §128 The rt. A ABU, ABC have AB common, and BE > BC. .■.Ai;>AC(% 101), and hence AE>AD. q.e.d. 515. CoE. 1. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpen- dicular ; and of two unequal lines the greater meets the plane at the greater distance from the foot of the perpendicular. 516. Cor. 2. The locus of a point in space equidistant from all points in the circumference of a circle is a straight line through the centre, perpendicular to the plane of the circle. LINES AND PLANES. 259 517. Cor 3. The locus of a point in space equidistant from the extremities of a straight line is the plane perpen- dicular to this line at its middle point. For any point C in this plane lies in a _L to AB at 0, its middle point; hence, CA and CB are equal. § 160 And any point D without the plane MH cannot lie in a _L to AB at 0, and hence is unequally distant from A and £. § 160 Pkoposition VI. Theorem. 518. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line in the plane, the line drawn from its intersection loith the line in the plane to any point of the perpendicular is per- pendicular to the line' of the plane. Let AB be a perpendicular to the plane MN, BE a perpendicular from B to any line CD in MN, and EA any line from E to AB. To prove that AE is ± to CD. Proof. Take EC and ED equal ; draw BC, BD, AC, AD. Now BC = BD. § 95 .-. AC = AD. § 514 .-. AE is ± to CD. § 161 ■ Q.E.D. 260 BOOK VI. SOLID GEOMETRY. Proposition VII. Theorem. 519. Two straight lines perpendicular to the same plane are parallel. ^N. ■--. .F \ Let AB be perpendicular to MN at B, and CD to MN at D. To prove that AB and CD are parallel. Proof, rrom A, any external point in AB, draw AD. Draw BD, and tlirougli D, draw UF in the plane MN± to BD. Then CD is ± to Bf' (§ 501), and AD is J. to BF. § 518 Therefore, CD, AD, and BD lie in the same plane. § 608 Also the line AB lies in this plane. § 492 But AB and CD are both _L to BD. § 501 Therefore, AB and CD are parallel. § 104 Q.E.D. 520. CoE. 1. If one of two 2}arallel lines is perpendicular to a plane, the other is also perpendicular to the plane. For if through any point O of CD a line is ^ c drawn ± to ilfiV", it is II to AB (§ 519),- and CD j^^ coincides with this ± and is X to MN. § 105 Eza 521. CoE. 2. If two straight lines are par- allel to a third straight line, they are par- allel to each other. to CD is J_ to AB and § 520 For a plane MN EF. M\ A E V B D M 'N LINES AND PLANES. 261 Peoposition VIII. Theorem. 522. If two straight lines are parallel, every plane containing one of the lines, and only one, is parallel to the other line. Let AB and CD ie two parallel lines, and MN any plane contain- ing CO but not AB. To prove that AB and MN are parallel. Proof. The lines AB and CD are in the same plane, § 103 and this plane intersects the plane MN in the line CD. Hyp. Therefore, if AB meets the plane MN at all, the point of meeting must be in the line CD. But since AB is II to CD, AB cannot meet CD. Therefore, AB cannot meet the plane MN. Hence, AB is II to MN. § 502 Q.E.D. 523. CoE. 1. Through either of two straight lines not in the same plane one plane, and only one, can he passed parallel to the other. For if AB and CD are the lines, and we ^ pass a plane through CD and the line CE drawn II to AB, the plane MN determined by CD and CE is II to AB. § 522 — B 262 BOOK VI. SOLID GEOMETRY. 524. CoE. 2. Through a given point a plane can be passed parallel to any two given straight lines in space. ITor if is the given point, and AB and n, CD the given, lines, by dravsring through a/\ a line A'B' II to AB, and also a line CD' II to CD, we shall have two lines A'B^ and CD' which determine a plane passing through and II to each of the lines AB and CD. § 522 N Proposition IX. Theorem. 525. If a straight line is parallel to a plane, the inter- section of the p)lane loith any plane passed through the given line is parallel to that line. Let the line AB be parallel to the plane MN, and let CD be the intersection of MN with any plane AD passed through AB. To prove that AB and CD are parallel. Proof. The lines AB and CD are in the same plane AD. Since CD lies in the plane JfiV", if AB meets CD it must meet the plane MN. But AB is by hypothesis II to MN, and therefore cannot meet it ; that is, it cannot meet CD, however far they may be produced. Hence, AB and CD are parallel. § 103 Q. E. D. LINES AND PLANES. 263 526. Cob. If a given straight line and a plane are parallel, a parallel to the given line drawn through any point of the plane lies in the plane. For the plane determined by the given line AB and any point C of the plane cuts MN in a line CD II to AB (§ 525) j but through C only one parallel to AB can be drawn (§ 105) ; therefore, a line drawn through C II to AB coincides with CD, and hence lies in the plane MN. Peopositiojst X. Theorem. 527. Two planes perpendicular to the same straight line are parallel. Let MN and PQ be two planes perpendicular to the straight line AB. To prove that MN and PQ are parallel. Proof. MN and PQ cannot meet. For if they could meet, we should have two planes from a point of their intersection J- to the same straight line. But this is impossible. § 510 Therefore, MN and PQ are parallel. § 604 Q.E.D. Ex. 604. Find the loou.s of points in space equidistant from two given parallel planes. Ex. 605. Eind the locus of points in space equidistant from two given points and also equidistant from two given parallel planes. 264 BOOK VI. SOLID GEOMETRY. Peoposition XL Thboebm. 528. The intersections of two parallel planes hy a third plane are parallel lines. Let the parallel planes MN and PQ be cut by RS. To prove that the intersections AB and CD are parallel. Proof. AB and CD are in the same plane RS. They are also in the parallel planes MN and PQ, which cannot meet, however far they extend. § 604 Therefore, AB and CD cannot meet, and are parallel. § 103 Q. E. D. 529. CoK. 1. Parallel lines included between parallel planes are equal. Eor if the lines A C and BD are parallel, the plane of these lines will intersect MN and PQ in the parallel lines AB and CD. § 628 .-.ABDC is a parallelogram. § 16C .-.AC and BD are equal. § 178 LINES AND PLANES. 265 530. Cor. 2. Two 'parallel planes are everywhere equally distant. For _k dropped from any points in MN to PQ measure the distances of these points from PQ. But these Js are parallel (§ 619), and hence equal (§ 629). Therefore, all points in MN are equidistant from PQ. Proposition XII. Theorem. 531. A straight line perpendicular to one of two parallel planes is perpendicular to the other also. 1 i ^-.—- A_ ,C 1 \. 1 \ ! 1 f"- — B j 1 — \. — '■ I Let AB be perpendicular to MN and PQ parallel to MN. To prove that AB is perpendicular to PQ. Proof. Pass through the line AB any two planes intersect- ing MJSr in the lines AC and AD, and PQ in BE and BF. Then AC and AD are II to BE and BF, respectively. § 628 But .4^ is _L to ^C and AD. § 601 .-. ^5 is -L to their parallels BE and BF. § 107 Therefore, AB ia A- to PQ. § 607 Q.E.D. 532. CoE. Through a given point one plane, and only one, can he drawn parallel to a given plane. Por if a line is drawn from ^ _L to PQ, a plane passing through ^ _L to this line is W to PQ (§ 627) ; and since through a point in a line only one plane can be drawn _L to the line (§ 609), only one plane can be drawn through A II to PQ. 266 BOOK VI. SOLID GEOMETRY. Proposition XIII. Theorem. 533. Iftioo intersecting straight lines are each parallel to a plane, the plane of these lines is parallel to that plane. M , _ f"" B N ^Q Let AC and AD be each parallel to the plane PQ, and let MN be the plane passed through AC and AD. To prove that MN is parallel to PQ. Proof. Draw AB ± to FQ. Pass a plane througla AB and AC intersecting PQ in BE, and a plane through AB and AD intersecting PQ in BF. Then AB is ± to BE and BF. § 501 Also, BE is II to AC, and BF is II to AD. § 525 Therefore, ^_B is ± to ^C and to AD. § 107 Therefore, AB is ± to the plane MN. § 507 Hence, MN and PQ are parallel. § 527 Q.E.D. Ex. 606. Find the locus of all lines drawn through a given point, parallel to a given plane. Ex. 607. Find the locus of points in a given plane which are equi- distant from two given points not in the plane. Ex. 608. Find the locus of a point in space equidistant from three given points not in a straight line. Ex. 609. Find a point in a plane such that the sum of its distances from two given points on the same side of the plane shall be a minimum. LINES AND PLANES. 267 Proposition XIV. Theorem. 534. If two angles not in the same plane have their sides respectively parallel and lying on the same side of the straight line joining their vertices, they are equal, and their planes are parallel. n T* ! J 'D \ 'N Q Let the corresponding sides of angles A and A' in the planes MN and PQ be parallel, and lie on the same side of AA'. To prove that Z.A = Z.A\ and that MN is II to PQ. Proof. Take AD and A'B' equal, also AC and A'C" equal. Draw DD', CC, CD, CD'. Since AD is equal and II to A'D', the figure ADDA' is a parallelogram, and AA' is equal and II to DD'. § 183 In like manner AA' is equal and II to CC. Also, since CC and DD' are eacli II to AA', and equal to AA', they are II and equal. .■.CD= CD'. .-.AADC = AA'D'C. . .ZA = ZA'. Now PQ is il to each of the lines AC and AD. Therefore, PQ is II to MN, the plane of these lines. §183 §160 §128 § 622 § 633 Q.E.D. 268 BOOK VI. SOLID GEOMETRY. Proposition XV. Theorem. 535, If two straight lines are cut by three parallel planes, their corresponding segments are proportio^ial. ^1 TT- / ^rr \ 1 ISA \ Let AB and CD be intersected by the parallel planes MN, PQ, RS, in the points A, E, B, and C, F, D. To prove that AE -.EB = CF: FD. Proof. Draw AD cutting the plane PQ in. G. Pass planes through AB and AD, AD and DC, cutting MN, PQ, RS in AC, EG, GF, BD. Then ^(? is II to 52), and Gi?" is II to .J C. §528 .-. AE : EB = AG : GD, § 342 and CF : FD = AG : GD. .■.AE:EB=CF:FD. Ax. 1 Q.E.D, Ex. 610. The line AB meets tliree parallel planes in the points A, E, B; and the line CD meets the same planes in the points C, F, I). If AE = 6 inches, BE = 8 inches, CD = 12 inches, compute CF and FD. Ex. 611. To draw a perpendicular to a given plane from a given point "without the plane. Ex. 612. To erect a perpendicular to a given plane at a given point in the plane. DIHEDRAL ANGLES. 269 DIHEDRAL ANGLES. 536. Def. The opening between two intersecting planes is called a dihedral angle. 537. Def. The line of intersection AB of the planes is the edge, the planes MA and NB are the faces, of the dihedral angle. 538. A dihedral angle is designated by its edge, or by its two faces and its edge. Thus, the dihedral angle in the margin may be designated by AB, or by M-AB-N. 539. In order to have a clear notion of the magni- tude of the dihedral angle M-AB-N, suppose a plane at first in coincidence with the plane MA to turn about the edge AB, as indicated by the arrow, imtil it coincides with the plane NB. The magnitude of the dihedral angle M-AB-N is proportional to the amount of rota- tion of this plane. 540. Def. Two dihedral angles M-AB-N and P-AB-N are adjacent if they have a comm.on edge AB, and a common face NB, be- tween them. 541. Def. When a plane meets another plane and makes the ad- jacent dihedral angles equal, each of these angles is called a right dihedral angle. 542. Def. A plane is perpendicular to another plane if it forms with this second plane a right dihedral angle. 270 BOOK VI. SOLID GEOMETRY. 543. Dep. Two vertical dihedral angles are dihedral angles that have the same edge and the faces of the one are the prolongations of the faces of the other. 544. Def. Dihedral angles axe acute, obtuse, complementary, supplementary, under the same conditions as plane angles. 545. Dep. The plane angle of a dihedral angle is the plane angle formed by two straight lines, one in each plane, perpen- dicular to the edge at the same point. 546. Cob, The plane angle of a dihedral angle has the same magnitude from whatever point in the edge the perpendiculars are drawn. For any two such angles, as CAD, GIH, have their sides respectively parallel (§ 104), and hence are equal. § 534 547. The demonstrations of many properties of dihedral angles are identically the same as the demonstrations of analo- gous theorems of plane angles. The following are examples : 1. If a plane meets another plane, it forms with it two adjacent dihedral angles whose sum is equal to two right dihedral angles. 2. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are in the same plane. 3. If two planes intersect each other, their vertical dihedral angles are equal. 4. If a plane intersects two parallel planes, the alternate- interior dihedral angles are equal ; the exterior-interior dihe- dral angles are equal ; the two interior dihedral angles on the same side of the transverse plane are supplementary. DIHEDRAL ANGLES. 271 5. When two planes are cut by a third plane, if the alternate- interior dihedral angles are -equal, or the exterior-interior di- hedral angles are equal, and the edges of the dihedral angles thus formed are parallel, the two planes are parallel. 6. Two dihedral angles whose faces are parallel each to each are either equal or supplementary. Proposition XVI. Thboeem. 548. Two dihedral angles are equal if their plane angles are equal. E, ,0 E', '-^c' Let the two plane angles ABD and A'B'D' of the two dihedral angles D-CB-E and D'-C'B'-E' be equal. To prove the dihedral angles D-CB-E and D'-C'B'-E' equal. Proof. Apply D'-C'B'-E' to D-CB-E, making the plane angle A'B'D' coincide with its equal ABD. The line B'C being ± to the plane A'B'D' will likewise be J- to the plane ABD at B, and fall on BC, since at B only one _L can be erected to this plane. § 511 The two planes A' B'C and ABC, having in common two intersecting lines AB and BC, coincide. § 497 In like manner the planes D'B'C and DBC coincide. Therefore, the two dihedral angles coincide and are equal. Q.E.D 272 BOOK VI. SOLID GEOMETRY. Pboposition XVII. Theorem. 549. Two dihedral angles have the same ratio as their plane angles. Fig. 2. Fig. 3. Fig. 1. Let A-BC-D and A'-B'C'-D' be two dihedral angles, and let their plane angles be ABD and A'B'D', respectively. To prove that A'-B'C'-D' : A-BC-D = Z A'B'D' : Z ABD. (Iase 1. When the plane angles are commensurable. Proof. Suppose the A ABD and A'B'D' (Figs. 1 and 2) have a common measure, which is contained ?« times in Z ABD and n times in Z A'B'D'. Then Z A'B'D' -.Z ABD = n:m. Apply this measure to Z ABD and Z A'B'D', and through the lines of division and the edges BC and B'C pass planes. These planes divide A-BC-D into m parts, and A'-B'C'-D' into n parts, equal each to each. § 548 Therefore, A'-B' C'-D' : A-B C-D = n: m. Therefore, A'-B'C'-D' : A-BC-D = Z A'B'D' : Z ABD. Ax. 1 Case 2. When the plane angles are incommensurable. Proof. Divide the Z ABD into any number of equal parts, and apply one of these parts to the Z A'B'D' (Figs. 1 and 3) as a unit of measure. DIHEDRAL ANGLES. 273 Since Z ABD and Z. A'B'D' are incommensurable, a certain number of these parts will form the Z A'B'E, leaving a re- mainder Z EB'D', less than one of the parts. Pass a plane through B'E and B'C. Since the plane angles of the dihedral angles A-BC-D and A'-B'C'-E ai'e commensurable, A'-B'C'-E : A-B C-D = Z A'B'E : /l ABD. Case 1 By increasing the number of equal parts into which Z ABD is divided, we can diminish at pleasure the magnitude of each pai't, and therefore make Z EB'D' less than any assigned value, however small, since Z EB'D' is always less than one of the equal parts into which Z ABD is divided. But we cannot make Z EB'D' equal to zero, since by hy- pothesis Z ABD and Z A'B'D' are incommensurable. § 269 Therefore, Z EB'D' approaches zero as a limit, if the num- ber of parts into which Z ABD is divided is indefinitely increased; and the corresponding dihedral angle E-B'C'-D' approaches zero as a limit. § 276 Therefore, Z A'B'E approaches Z A'B'D' as a limit, § 271 and A'-B'C'-E approaches A'-B'C'-D' as a limit. Z. A'B'E Z A'B'D' Hence, , approaches / AT>r, ^ ^ limit, § 280 A'-B'C'-E , A'- B'C'-D' ,. .^ „ ^^^ and , „^ .^ approaches . „^ -_ as a limit. § z80 A-BC-D ^ A-BC-D ^ ^ Z A'B'E . ^ ^, , , A'-B'C'-E „ . But , IS -constantly equal to > Case 1 as Z EB'D' varies in value and approaches zero as a limit. Therefore, the limits of these variables are equal. § 284 That is A'-B'C'-D' _ Z A'B'D' _ mat IS, A-BC-D ~ ZABD q.e.d. 550, CoK. The plane angle of a dihedral angle may he taken as the measure of the dihedral angle. 274 BOOK VI. SOLID GEOMBTUY. Proposition XVIII. Theorem. 551. If tivo planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other plane. Let the plane PQ be perpendicular to MN, and let CD be drawn in PQ perpendicular to AB, the intersection of PQ and MN. To prove that CD is perpendicular to MN. Proof. In the plane MN draw DE _L to AB at D. Then CDE is the measure of the right dihedral angle P-AB-N, and is therefore a right angle. § 550 But, by hypothesis, CD A is a right angle. Therefore, CD is _L to DA antj DE at their point of inter- section, and consequently X to their plane MN. § 507 Q.E.D. 552. CoR. 1. If two planes are perpendicular to each other, a perpendicular to one of them at any point of their inter- section will lie in the other plane. For a line CD drawn in the plane PAB _L to AB at the point D will be X to MN {^ 651). But at the point D only one A- can be drawn to MN {% 511). Therefore, a ± to MN erected at D will coincide with CD and lie in the plane PAB. DIHEDRAL ANGLES. 275 553. Cor. 2. If two planes are perpendicular to each other, a perpendicular to one of them from any point of the other will lie in the other plane. For a line CD drawn in the plane PAB from the point C _L to AB will be _L to MN (§ 551). But from the point C only one ± can be drawn to MN (§ 511). Therefore, a ± to MN drawn from C will coincide with CD and lie in PAB. Pboposition XIX. Theorem. 554. If a straight line is perpendicular to a plane, every plane passed through this line is perpendicular to the plane. Let CD be perpendicular to MN, and PQ be any plane passed through CD intersecting MN in AB. To prove that PQ is perpendicular to the jjlane MN Proof. Draw DE in the plane MN X to AB. Since CD is ± to MN, it ia 1-to AB. § 501 Therefore, Z CDE is the measure of P-AB-N § 550 But Z CDE is a right angle. § 501 Therefore, PC> is ± to MN. § 542 Q. E. D. 555. Cor. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of its faces. 276 BOOK VI. SOLID GE'OMETRY. Proposition XX. Theorem. 556. If two intersecting planes are each 2^er2oendicular to a third plane, their intersection is also perpendicular to that plane. Let the planes BD and BC intersecting in the line AB be perpen- dicular to the plane PQ. To prove that AB is perpendicular to the plane PQ. Proof. A X erected to P() at -B, a point cominon to the three planes, will lie in the two planes BC and BD. § 652 And since this ± lies in both the planes BC and BD, it must coincide with their intersection AB. § 606 .-. AB is J_ to the plane PQ. Q.E.D. 557. CoR. 1. If a plane is perpendicular to each of two intersecting planes, it is perpendicular to their intersection. 558. Cor. 2. If a plane is perpendicular to each of two planes that include a right dihedral angle, the intersection of any two of these planes is perpendicular to the^third plane, and each of the three intersections is perpendicular to the other two. DIHEDRAL ANGLES. 277 Proposition XXI. Theorem. 559. Every point m a plane ivhich bisects a dihedral angle is equidistant from the faces of the angle. Let the plane AM bisect the dihedral angle formed by the planes AD and AC ; and let PE and PF be perpendiculars drawn from any point P in the plane AM to the planes AC and AD. To prove that PE = PF. Proof. Through PE and PF pass a plane intersecting the planes AC, AD, and AM in the lines OF, OF, and PO. TheplaneP^i^ is J-to^C and to JZ». §554 Hence, the plane PEF is J_ to their intersection AO. % 557 .-. .40 is J_ to OE, OP, and OF. .-. Z PO^ is the measure of M-AO-G, and A P OF is the measure of M-A 0-D. But M-A 0-0 = M-A 0-D. .-.^POE = APOF. .-. rt. A POE = rt. A POF. . . PE = PF. §501 § 550 Hyp. §141 § 128 Q. E. D. 278 BOOK VI. SOLID GEOMETRY. Proposition XXII. Theorem. 560. Through a given straight line not perpendicular to a plane, one plane, and only one, can he passed per- pendicular to the given plane. Let AB be the given line not perpendicular to the plane MN. To prove that one plane can he passed through AB perpen- dicular to MN, and only one. Proof. From any point B of AB draw UK _L to MN, and through AB and HK pass a plane AF. The plane AF is ± to MN, since it passes through HK, a line _L to MN. § 664 Moreover, if two planes could be passed through AB _L to the plane MN, their intersection AB would be _L to MN. § 656 But this is impossible, since AB is by hypothesis not per- pendicular to the plane MN. Hence, through AB only one plane can be passed ± to MN. 561. Def. The projection of a point on a plane is the foot of the perpendicular from the point to the plane. 562. Def. The projection of a line on a plane is the locus of the projections of its points on the plane. DIHEDEAL ANGLES. 279 Peoposition XXIII. Theorem. 563, The projection of a straight line not perpendicu- lar to a plane upon that plane is a straight line. Let AB be the given line, MN the given plane, and CD the projec- tion of AB upon MW. To prove that CD is a straight line. Proof. From any point H of AB draw HE 1. to MN, and pass a plane AF through HK and AB. The plane AF is ± to MN, § 554 and contains all the Js drawn from AB to MN. § 553 Hence, CD must be the intersection of these two planes. Therefore, CD is a straight line. § 506 Q.E. D. 564. CoE. The projection of a straight line perpendicular to a plane upon that plane is a point. 565. Dee. The plane ^i? CD is called the projecting plane of the line AB upon the plane MN. 566. Dee. The angle which a line makes with a plane is the angle which it makes with its projection on the plane ; and is called the inclination of the line to the plane. 280 BOOK VI. SOLID GEOMETRY. Proposition XXIV. Theorem. 567. The acute angle ivhich a straight line makes with its projection upon a plane is the least angle which it makes with any line of the plane. Let BA meet the plane MN at A, and let AC be its projection upon the plane MN, and AD any other line drawn through A in the plane. To prove that Z BAG is less than Z BAD. Proof. Take AD equal to AC, and draw BD. In the ^BAC and BAD, BA = BA, Iden. AC = At), Const. but BC C. .-. C'C is J_ to ^5 and X>C. Again, C'C is the only _L to both AB and DC. For, if pos- sible, let EA be any other line J_ to AB and DC. Then EA is ± to ^(? (§ 107), and hence _L to MN. § 507 Draw EH ± to D'C. Then EHi& _L to MN (§ 551), and we have two Js from E to MN. But this is impossible. § 511 Hence, C'C is the only common J_ to i>6' and AB. q. e.d. § 601 § 501 §107 282 BOOK VI. SOLID GEOMETRY. POLYHEDRAL ANGLES. 569. Dbf. The opening of three or more planes which meet at a common point is called a polyhedral angle. 570. Def. The common point S is the vertex of the angle, and the intersections of the planes SA, SB, etc., g are its edges ; the portions of the planes included A between the edges are its faces, and the angles /V \ formed by the edges are its face angles. '^/--/- \ 571. The magnitude of a polyhedral angle de- jB pends upon the relative position of its faces, and not upon their extent. 572. In a polyhedral angle, every two adjacent edges form a face angle, and every two adjacent faces form a dihedral angle. These face angles and dihedral angles are the parts of the polyhedral angle. 573. Def. A polyhedral angle is convex, if every section made by a plane that cuts all its edges is a convex polygon. 574. Def. A polyhedral angle is called trihedral, tetrahedral, etc., according as it has three faces, four faces, etc. 575. Def. A trihedral angle is called rectangular, bi-rectan- gular, tri-rectangular, according as it has ojie, two, or three right dihedral angles. 576. Def. A trihedral angle is called isosceles if it has two of its face angles equal. g „, 577. Def. Two polyhedral angles yf\ /f\ can be made to coincide and are equal // \ // \ if their corresponding parts are equal A'/-~- \ y^"/-~ \ and arranged in the same order. /b ^ 'b' \ 578. A polyhedral angle is designated by its vertex, or by its vertex and all the faces taken in order. _Thus the poly- POLYHEDRAL ANGLES. 283 hedial angle in the margin may be designated by (S*, or by S-ABCD. 579. If the faces of a polyhedral angle S-ABGD are produced through the vertex S, another polyhedral angle S-A'B'C'D' is formed, sym- metrical with respect to S-ABCD. The face angles ASB, BSC, etc., are equal, re- spectively, to the face angles A' SB', B'SC, etc. § 93 Also the dihedral angles SA, SB, etc., are equal, re- spectively, to the dihedral angles SA', SB', etc. § 647 (The second figure shows a pair of vertical dihedral angles.) The edges of S-ABCD are ai-ranged from left to right (counter clockwise) in the order SA, SB, SC, SB, but the edges of S-A'B'C'D' are arranged from right to left (clockwise) in the order SA\ SB', SC, SD' ; that is, in an order the reverse of the order of the edges in S-ABCD. Two symmetrical polyhedral angles, therefore, have all their parts equal, each to each, but arranged in reverse order. In general, two symmetrical polyhedral angles are not super- posable. Thus, if the trihedral angle S-A'B'C is made to turn 180° about XY, the bisector of the angle A'SC, then SA' wiU coincide with SC, SC with SA, and the face A'SC with ASC; but the dihedral angle SA, and hence the dihedral angle SA', not being equal to SC, the plane A' SB will not coincide with BSC ; and, for a similar reason, the plane C'SB' will not coincide with ASB. Hence, the edge SB' takes some position SB" not coinci- dent with SB ; that is^ the trihedral angles are not superposable. 284 BOOK VI. SOLID GEOMETRY. Peoposition XXVI. Theorem. 580. The sum of any two face angles of a trihedral angle is greater than the third face angle. In the trihedral angle S-ABC, let the angle ASC be greater than ASB or BSC. To prove Z ASB + Z BSC greater than Z ASC. Proof. In ASC draw SD, making Z ASD equal to Z ASB. Througli any point D of SD draw ADC in the plane ASC. Take SB equal to SD. Pass a plane through the line AC and the point B. The A ASD and ASB are equal. § 143 Por AS = AS, SD = SB, and Z ASD = Z .46'5. .■.AD = AB. §128 In the A .45 C, But By subtraction, AB + BO AC. AB = AD. §138 Ax. 6 5a>z>c. In the A ^-SC and D^S'C, SC = SC, and 5.B = SD, but BODC. Therefore, Z 55'C is greater than Z Z»^C. § 166 .-. Z -4,S'5 + Z BSC is greater than Z ASD + Z D/SC. That is, Z ^^S'.B + Z BSC is greater than Z ASC. q.e.d. POLYHEDRAL ANGLES. 285 Proposition XXVII. Theorem. 581. The sum of the face angles of any convex poly- hedral angle is less than four right angles. Let S be a convex polyhedral angle, and let all its edges be cut by a plane, making the section ABODE. To prove Z. ASB + Z. BSC, etc., less than four rt. A. Proof. From any point within the polygon draw OA, OB, OC, OD, OK The number of the A having the common vertex is the same as the number having the common vertex S. Therefore, the sum of the A of all the A having the common vertex S is equal to the sum of the A of aU the A having the common vertex 0. But in the trihedral A formed at A, B, C, etc., Z SAE + Z SAB is greater than Z EAB, Z SB A + ASBCis greater than Z ABC, etc. § 580 Hence, the sum of the A at the bases of the A whose com- mon vertex is ;S^ is greater than the sum of the A at the bases of the A whose common vertex is 0. Ax. 4 Therefore, the sum of the A at the vertex S is less than the sum of the A at the vertex 0. Ax. 5 But the sum of the Z at is equal to 4 rt. A. § 88 Therefore, the sum of the Z at ^ is less than 4 rt. A. Q. E. D. 286 BOOK VI. SOLID GEOMETRY. Proposition XXVIII. Theorem. < 582. T^oo trihedral angles are equal or symmetrical ivhen the three face angles of the one are respectively equal to the three face angles of the other. B' E' A' B E A a! E B' In the trihedral angles S and S', let the angles ASB, ASC, BSC be equal to the angles A'S'B', A'S'C, B'S'C, respectively. To prove that S and S' are equal or symmetrica!. Proof. On the edges of these angles take the six equal dis- tances SA, SB, SC, S'A<, S'B', S'C. Draw AB, BC, AC, A'B', B'C, A'C. The isosceles A SAB, SAC, SBC are equal, respectivelj', to the isosceles A S'A'B', S'A'C, S'B'C. § 143 .-. AB, BC, CA are equal, respectively, to A'B', B'C, C'A'. .■.AABC = AA'B'C'. §150 At any point D in SA draw DE in the face ASB and DF in the face .J,S'C-Lto SA. These lines meet AB and AC, respectively, {since the A SAB and SAC are acute, each being one of the equal A of an isosceles A). Draw EF. On A'S' take A'D' equal to AJ). §142 Const. § 128 §128 § 143 § 128 § 128 §150 POLYHEDRAL ANGLES. 287 Draw D'E' in the face A'S'B' and D'F' in the face A'S'C ± to S'A', and draw Ei^ = D'F'. .■.AAEF = AA'E'F'. For ^^ = A'E', AF== A'F', and Z ^4i^ = Z ^'^'i^'. .-. EF = E'F'^ .■.AEDF = AE'D'F'^ For ^J* = E'D', DF = D'i?", and EF = ^'i?". . . Z ^i>i^ = Z E'D'F'. § 128 .-. the angle B-AS-C = B'-A'S'-C, {since A EBF and E'D'F', the measures of these dihedral A, are equal). In like manner it may be proved that the dihedral angles A-BS-G and A-CS-B are equal, respectively, to the dihedral angles A'-B'S'-C and A'-C'S'-B'. .'. S and S' are equal or symmetrical. §§ 577, 679 Q.E.D. This demonstration applies to either of the two figures de- noted by S'-A'B'C, which are symmetrical with respect to each other. If the first of these figures is taken, S and S' are equal. If the second is taken, S and S' are symmetrical. 583. Cob. If two trihedral angles have three face angles of the one equal to three face angles of the other, then the dihedral angles of the one are respectively/ equal to the dihe- dral angles of the other. 288 BOOK VI. SOLID GEOMETRY. EXERCISES. Ex. 616. rind the locus of points in space equidistant from two given intersecting lines. Ex. 617. Find the locus of points in space equidistant from all points in the circumference of a circle. Ex. 618. Find the locus of points in a plane equidistant from a given point without the plane. Ex. 619. Find a point at equal distances from four points not all in the same plane. Ex. 620. Two dihedral angles which have their edges parallel and their faces perpendicular are equal or supplementary. Ex. 621. The projections on a plane of equal and parallel lines are equal and parallel. Ex. 622. Two trihedral angles are equal when two dihedral angles and the included face angle of the one are equal, resj)ectively, to two dihedral angles and the included face angle of the other, and similarly placed. Ex. 623. Two trihedral angles are equal when two face angles and the included dihedral angle of the one are equal, respectively, to two face angles and the included dihedral angle of the other, and similarly placed. Ex. 624. If the face angle ASB of the trihedral angle S-ABC is bisected by the line SD, the angle CSD is less than, equal to, or greater than half the sum of the angles ASC and BSC according as CSD is less than, equal to, or greater than a right angle. Ex. 625. An isosceles trihedral angle and Its symmetrical trihedral angle are superposable. Ex. 626. Find the locus of points equidistant from the three edges of a trihedral angle. Ex. 627. Find the locus of points equidistant from the three faces of a trihedral angle. Ex. 628. If two face angles of a trihedral angle are equal, the dihedral angles opposite them are equal. Ex. 629. The planes that bisect the dihedral angles of a trihedral angle intersect in the same straight line. Book VII. POLYHEDEONS, CYLINDERS, AND CONES. POLYHEDRONS. 584. Def. a polyhedron is a solid bounded by planes. The bounding planes are called the faces, the intersections of the faces, the edges, and the intersections of the edges, the vertices, of the polyhedron. 585. Def. A diagonal of a polyhedron is a straight line join- ing any two vertices not in tlie same face. 586. Def. A section of a polyhedron is the figure formed by its intersection with a plane passing through it. 587. Def. A polyhedron is convex if every section is a con- vex polygon. Only convex polyhedrons are considered in this work. iiii| 1 Tetrahedron. Hexahedron. Octahedron. Dodecahedron. Icosahedron. 588. Def. A polyhedron of four faces is called a tetrahe- dron ; one of six faces, a hexahedron ; one of eight faces, an octahedron ; one of twelve faces, a dodecahedron ; one of twenty faces, an icosahedron. Note. Full lines in the figures of solids represent visible lines, dashed lines represent invisible lines. 289 290 BOOK VII. SOLID GEOMETRY. PRISMS AND PARALLELOPIPEDS. 589. Def. a prism is a polyhedron of which two faces are equal j)olygoris in parallel planes, and the other faces are parallelograms. The equal ijolygons are called the bases of the prism, the parallelograms, the lat- eral faces, and the intersections of the lat- eral faces, the lateral edges of the prism. The sum of the areas of the lateral faces of a prism is called its lateral area. Prism. Eight Prism. 590. Def. The altitude of a prism is the perpendicular distance bet'ween the planes of its bases. 591. Def. A right prism is a prism whose lateral edges are perpendicular to its bases. 592. Def. A regular prism is a right j.irism wh(ise bases are regular polj'gons. 593. Def. An oblique prism is a prism whose lateral edges are oblique to its bases. 594. The lateral edges of a prism are equal. The lateral edges of a right prism are equal to the altitude. 595. Def. Prisms are called triangular, quadrangular, etc., according as their bases are triangles, quadrilaterals, etc. Triangular Prism. 596. Def. A parallelopiped is a prism whose bases are paral- lelograms. PRISMS AND PARALLELOPIPEDS. 291 Kectangular Parallelopiped. Cube. Oblique Parallelopiped. 597. Def. a right parallelopiped is a parallelopiped whose lateral edges are perpendicular to the bases. 598. Def. a rectangular parallelo- piped is a parallelopiped whose six faces are all rectangles. 599. Def. A cube is a parallelo- — ' piped whose six faces are all squares. - 600. Def. A cube whose edges are equal to the linear unit is taken as the Riglit Section of a Prism. unit of VOlume. 601. Def. The volume of any solid is the number of units of volume which it contains. 602. Def. Two solids are equivalent if their volumes are equal. 603. Def. A right section of a prism is a section made l)v a plane perpen- dicular to the lateral edges of the prism. 604. Def. A truncated prism is the part of a prism included between the base and a section made by a plane oblique to the base. Truncated Prism. 292 BOOK VII. SOLID GEOMETRY. Proposition I. Theobem. 605. The sections of a prism made hy parallel planes cutting all tlie lateral edges are equal polygons. Let the prism AD be intersected by parallel planes cutting all the lateral edges, making the sections GK, G'K'. To prove that GK = G'K'. Proof. The sides GH, HI, IK. etc., are parallel, respectively, to the sides G'H', IJ'I', I'K. etc. § 528 The sides GH, HI, IK, etc., are equal, respectively, to G'H'. H'l', I'K', etc. § 180 The A GHI. HIK, etc., are equal, respectively, to A G'H'I'. H'I'K'. etc. § 534 Therefore, GK= G'K' § 203 Q.E.D. 606. CoE. Every section of a prism made liy a plane parallel to the base is equal to tJie baxe ; and all right sections of a prism are equal. Ex. 630. The diagonals of a parallelepiped bisect one another. Ex. 631. The lateral faces of a right prism are rectangles. Ex. 632. Every section of a prism made by a plane parallel to the lateral edges is a parallelogram. PRISMS AND PARALLELOPIPEDS. 293 Proposition II. Theokem. 607. The lateral area of a prism is equal to the product of a lateral edge hy the perimeter of the right section. Let GHIKI, be a right section of the prism AD', S its lateral area, E a lateral edge, and P the perimeter of the right section. To frove that S = E X P. Proof. AA' = BB' = CC" = DD' • • • = ^. § 594 GH is ± to BB', HI to CC, IK to DD', etc. § C0.3 .-. the area of O AB' = BB' X GFI= Ex GH, § 400 the area of EJ BC = C'6" X HI = E X HI, and so on. Therefore, S, the sum of these parallehigrams, is equal to E{GII + HI+ IK + etc.). But GH + HI + IK + etc. = P. Therefore, S = E X P. q.e.d. 608. Con. The lateral area of a. right prism is equal to the product of the altitude hy the perimeter of the base. Ex. 633. Find the lateral area of a right prism, if its altitude is 18 inches and the perimeter of its base 29 inches. 294 BOOK VII. SOLID GEOMETRY. Proposition III. Theoeem. 609. Ttvo prisms are equal if the three faces which include a trihedral angle of the one are respectively equal to three faces which include a trihedral angle of the other, and are similarly placed. y FfC — T — -yf F' In the prisms AI and AT, let the faces AD, AG, AJ be respectively equal to A'D', A'G', A'J', and similariy placed. To prove that AI = A' I'. Proof. The face A BAE, BAF, EAF axe equal to the face A BA'jE', BA'F', FA'F, respectively. § 203 Therefore, the trihedral angles A and A' are equal. § 682 Apply the trihedral angle A to its equal A'. Then the face AD coincides -with A'D', AG -with A'G', and AJ with A'J; and C falls at C, and D at D'. The lateral edges of the prisms are parallel. § 589 Therefore, CIT falls along C'H', and DI along DT. § 105 Since the points F, G, and J" coincide with F', G', and J', each to each, the planes of the upper bases coincide. § 496 Hence, H coincides with H', and I with /'. Therefore, the prisms coincide and are equal. q.e.d. 610. CoE. 1. Two truncated prisms are equal under the hypothesis given in § 609. 611. CoE. 2. Two right prisms having equal bases and equal altitudes are equal. PRISMS AND PARALLELOPIPEDS. 295 Pkoposition IV. Theorem. 612. An oblique prism is equivalent to a right prism whose base is equal to a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Fir- Let FI be a right section of the oblique prism AD', and FI' a right prism whose lateral edges are equal to the lateral edges of AD'. To prove that AD' =0= FT. Proof. If from the equal lateral edges of AD' and FI' we take the lateral edges of FD', -which are cominon to both, the remainders AF and A'F', BG and B'G', etc., are equal. Ax. 3 The upper bases FI and F'l' are equal. § 589 Place AI on A'F so that FI shall coincide with F'l'. Th&n^FA, GB, etc., coincide with F'A', G'B', etc. § 511 Hence, the faces GA and G'A', BB and WB', coincide. But the faces FI and F'F coincid-e. .•. the truncated prisms AI and AT are equal. § 610 Now AI + FD' = AD', Ax. 9 and A'F + FD' = FF. Therefore, AD'=o=Fr. Ax. 2 Q. E. D. 296 BOOK VII. SOLID GEOMETRY. Pkoposition V. Theokem. 613. The opposite lateral faces of a parallelopiped are equal and parallel. Let BH be a parallelopiped with bases BD and FH. To prove that the opposite faces B G and AH are equal and parallel. Proof. BC ia equal and parallel to AD, §§ 178, 166 and BF is equal and parallel to AE. .■.AFBC = AEAD. §534 .-. BG iBWto AS. §534 .■.BG = AH. §185 In like manner it may be proved that the face BE is equal and parallel to CH. q.e.d. 614. CoE. Any tioo opposite faces of a parallelopiped mai/ be taken as bases. Ex. 634. Find the lateral area of a right prism, if its altitude is 20 inches, and its base is a triangle whose sides are 7 inches, 8 inches, and 9 inches, respectively. Ex. 635. Find the lateral area of a triangular prism, if its lateral edge is 20 inches, and its right section is a triangle whose sides are 9 inches, 10 inches, and 12 inches, respectively. Ex. 636. Find the total area of a right prism, if its altitude is 32 inches, and its base is a triangle whose sides are 12 inches, 14 inches, and 16 inches, respectively. PRISMS AND PARALLELOPIPEDS. 297 Proposition YI. Theorem. 615. The plane passed through two diagonally opjpo- site edges of a parallelopiped divides the parallelo2ri2Md into two equivalent triangidar prisms. Let the plane ACGE pass through the opposite edges AE and CG of the parallelopiped AG. To prove tliat the parallelopiped AG is divided info tico equivalent triangular prisms ABC-F and ADtJ-H. Proof. Let IJKL be a right section of the parallelopiped. The opposite faces AF and DG, and AH and BG, are parallel and equal. § 613 .•.i;7is II to iA', and 7X to ./A". §528 Therefore, IJKL is a pai'allelogram. § 166 The intersection IK of the right section with the plane ACGE is the diagonal of the O IJKL. . • . A UK = A ILK. § 179 But the prism ABC-F is ecjuivalent to a right prism whose base is UK and altitude AE, and the prism A CI)-H is equiva- lent to a right prism whose base is ILK, and altitude AE. § 612 But these two right prisms are equal. § 611 .-. ABC-F ^ ADC-H. Ax. 1 Q.E. D. 298 BOOK VII. SOLID GEOMETRY. Proposition VII. Theoebm. 616. Two rectangular parallelopipeds having equal bases are to each other as their altitudes. Let AB and A'B' be the altitudes of the two rectangular parallel opipeds P and P', which have equal bases. To prove that P:F' = AB: A'B'. Case 1. When AB and A'B' are commensurable. Proof. Find a common measure of AB and A'B'. Apply this common measure to A£ and A'B' as a unit of measui'e. Suppose this common measure to be contained m times in AB, and n times in A'B'. Then AB : A'B' = m:n. At the several points of division on AB and A'B' pass planes perpendicular to these lines. The parallelopiped P is divided into m parallelopipeds, and P' into n parallelopipeds, equal each to each. § 611 Therefore, P -.P' = m:n. Therefore, P:P' = AB: A'B'. Ax. 1 PRISMS AND PARALLELOPIPEDS. - 299 Case 2. When AB and A'H' are incommensurable. Proof. Let AB be divided into any number of equal parts, and let one of these parts be applied to A'B' as a unit of measure as many times as A'B' will contain it. Since AB and A'B' are incommensurable, a certain number of these parts will extend from A' to a point D, leaving a remainder I>B' less than one of the j)arts. Through D pass a plane ± to A'B', and let Q denote the parallelepiped whose base is the same as that of F', and whose altitude is A'B. Then Q-B = A'B : AB. Case 1 If the numlier of parts into which AB is divided is indefi- nitely increased, the ratio Q : P approaches P' : P as a limit, and the ratio A'D : AB approaches A'B' : AB as a limit. The theorem can be proved for this case by the Method of Limits in the manner shown in § 649. q.e.d. 617. Def. The three edges of a rectangular p)arallelopiped which meet at a common vertex are called its dimensions. 618. CoE. Tivo rectangular jmrallelopipeds ivhich have tivo dimensions in common are to each other as their third dimensions. 300 BOOK VII. SOLID GEOMETRY. Proposition VIII. Thboebm. 619. Two rectangular paralUlopipeds having equal altitudes are to each other as their hoses. let a, b, c, and a', b', c, be the three dimensions, respectively, of the two rectangular parallelopipeds P and P'. P axb P'~ a'Xb'' To prove that Proof. Let (? be a third rectangular parallelopiped whose dimensions are a', b, and c. Now Q has the two dimensions b and c in common with P, and the two dimensions a' and c in common with P'. Then P and Q = a'' Q P' b ~ b'' § 618 The products of the corresponding members of these two equalities give P^_ a X b P''^ a'Xb'' Q.E.D. 620. Cob. Two rectangular parallelopipeds which have one dimension in common are to each other as the products of their other tu>o dimensions. PEISMS AND PARALLELOPIPEDS. 301 Proposition IX. Theorem. 621. Two rectangular 2')arallelopipeds are to each other as the products of their three dimensions. Let a, b, c, and a', b', c', be the three dimensions, respectively, of the two rectangular parallelepipeds P and P'. To prove, that P a X li X P' a' X h' X c' Proof. Let Q- be a third rectangular paraUelopiped whose dimensions are a, V, and c. Then and p b b' Q n. X i; P' a' X o' 618 § 620 The products of the corresponding niembers of these equali- ties give P aXh X c. P' a' Xh' X o' Q.E.D. Ex. 637. Find the ratio of two rectangular parallelepipeds, if tlieir altitudes are each 6 inches, and their bases 5 inches by 4 inches, and 10 inches by 8 inches, respectively. Ex. 638. Find tlie ratio of two rectangular parallelopipeds, if their dimensions are 3, 4, 5, and 9, 8, 10, respectively. 302 BOOK VII. SOLID GEOMETRT. Proposition X. Theobem. 622. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. i'y y-'i ^ ^ E—--± (^ : -i Let a, b, and c be the three dimensions of the rectangular paral- lelopiped P, and let the cube U be the unit of volume. To prove that the volume of P = a X h Y. o. P a X b y. c Proof. U Ixlxl = a X b X 0. P . §621 Since U is the uiiit of volume, — is the volume of P. § 601 Therefore, the volume of P = a X b X c. q e d 623. CoE. 1. The volume of a cube is the cube of its edge. 624. Cor. 2. The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 625. Scholium. When the three dimensions of a rectan- gular parallelopiped are each exactly divisible by the linear unit, this proposition is rendered evident by dividing the solid into cubes, each equal to the unit of volume. Thus, if the three edges which meet at a common vertex contain the linear unit 3, 5, and 8 times respectively, planes passed through the several points of division of the edges, perpendicular to the edges, will divide the solid into 3x5x8 cubes, each equal to the unit of volume. PRISMS AND PARAXLELOPIPJnJS. 303 Peopositiojt XI. Theoebm. 626. The volume of any parallelopiped is equal to the product of its hase by its altitude. Let P be an oblique parallelopiped no two of whose faces are per- pendicular, whose base B is a rhomboid, and whose altitude is H. To prove that the volume of P = B % H. Proof. Prolong the edge EF and the edges II to EF, and cut them perpendicularly by two parallel planes whose distance apart GI is equal to FF. We then have the oblique parallelo- piped Q whose base C is a rectangle. Prolong the edge IK and the edges II to IK, and cut them perpendicularly by two planes whose distance apart MN is equal to IK. We then have the rectangular parallelopiped R. Now P =0= §, and ^ =0= ^. § 612 .-. P ^B. Ax. 1 The three solids have a common 'altitude II. § 530 Also B^C; §401 and C = Z>. § 186 .-. B ^B. Ax. 1 But the volume oi R = D X H. § 624 Putting P for R, and B for D, we have the volume oi. P = B X S. q. e. d. 30-4 BOOK VII. SOLID GEOMETRY. Pkoposition XII. Theorem. 627. The volume of a triangular prism is equal to the product of its hose by its altitude. Let V denote the volume, B the base, and H the altitude of the triangular prism CEA-E'. To prove that V = S X H. Proof. Upon the edges CJS, JSA, EE', construct the parallel- opiped CEAD-E'. Then CEA-E' ^ i CEAD-E. § 616 Now the voliune of CEAD-E' = CEAD X -H". § 626 But CEAD = 2B. § 179 .■.r = i(2Bxir) = BxJH:. . q.e.d. Ex. 639. Two triangular prisms are equal 1£ their lateral faces are equal, each, to each, and similarly placed. Ex. 640. The square of a diagonal of a rectangular paraUelopiped is equal to the sum of the squares of the three dimensions. Ex. 641. The sum of the squares of the four diagonals of a paraUelo- piped is equal to the sum of the squares of the twelve edges. Ex. 642. The volume of a triangular prism is equal to half the product of any lateral face hy the perpendicular dropped from the opposite edge on that face. PRISMS AND PARALLBLOPIPEDS. 305 Proposition XIII. Theorem. 628. The volume of any prism is equal to the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the prism DA'. To prove that V = B X M. Proof. Planes passed tkrough the lateral edge AA', and the diagonals AC, AD of the base, divide the given prism into triangular prisms that have the common altitude H. The volume of each triangular prism is equal to the product of its base by its altitude (§ 627) ; and hence the sum of the volumes of the triangular prisms is equal to the sum of their bases multiplied by their common altitude. But the sum of the triangular prisms is equal to the given prism, and the sum of their bases is equal to its base. Ax. 9 Therefore, the volume of the given prism is equal to the product of its base by its altitude. That is, V=BxH. q.e.d. 629. CoR. Two prisms are to each other as the products of their bases hy their altitudes ; prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes are to each other as their bases ; prisms having equivalent bases and equal altitudes are equivalent.^ 306 BOOK VII. SOLID GEOMETRY. PROBLEMS OF COMPUTATION. Ex. 643. If the edge of a cube is 15 inclies, find the area of the total surface of the cube. Ex. 644. If the length of a rectangular parallelepiped is 10 inches, its width 8 inches, and its height 6 inches, find the area of its total surface. Ex. 645. Find the volume of a right triangular prism, if .its height is 14 inches, and the sides of the base are 6, 5, and 5 inches. Ex. 646. The base of a right prism is a rhombus, one side of which is 10 inches, and the shorter diagonal is 12 inclies. The height of the prism is 15 inches. Find the entire surface and the volume. Ex. 647. Pind the volume of a regular prism whose height is 10 feet, if each side of its triangular base is 10 inches. Ex. 648. How many square feet of lead will be required to line a cis- tern, open at the top, which is 4 feet 6 inches long, 2 feet 8 inches wide, and contains 42 cubic feet ? Ex. 649. An open cistern 6 feet long and 4|- feet wide holds 108 cubic feet of water. How many square feet of lead will it take to line the sides and bottom ? Ex. 650. An open cistern is made of iron 2 inches thick. The inner dimensions are : length, 4 feet 6 inches ; breadth, 3 feet ; depth, 2 feet 6 inches. "What will the cistern weigh (i) when empty ? (ii) when full of water ? (The specific gravity of the iron is 7.2.) Ex. 651. Find the volume of a regular hexagonal prism, if its height is 10 feet, and each side of the hexagon is 10 inches. Ex. 652. Find the length of an edge of a cubical vessel that will hold 2 tons of water. Ex. 653. One edge of a cube is a. Find the surface, the volume, and the length of a diagonal of the cube. Ex. 654. A diagonal of one of the faces of a cube is a. Find the vol- ume of the cube. Ex. 655. The three dimensions of a rectangular parallelopiped are a, h, c. Find the area of its surface, its volume, and the length of a diagonal. Ex. 656. The volume of a parallelopiped is V, and the three dimen- sions are asm:n:p. Find the dimensions. PYRAMIDS. 307 /. // PYRAMIDS. 630. Def. a pyramid is a polyhedron of which, one face, called the base, is a polygon of any number of sides and the other faces are triangles hav- ing a common vertex. The faces which have a com- mon vertex are called the lat- eral faces of the p3'ramid, and their common vertex is called the vertex of the pyramid. Pyramids. 631. Def. The intersections of the lateral faces are called the lateral edges of the pyramid. 632. Def. The sum of the areas of the lateral faces is called the lateral area of the pyramid. 633. Def. The altitude of a pyramid is the length of the perpendicular let fall from the vertex to the plane of the base. 634. Def. A pyramid is called triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc. 635. Def. A triangular pyramid has four triangular faces, and is called a tetrahedron. Any one of its faces can be taken for its base. 636. Def. A pyramid is regular if its base is a regular polygon whose centre coincides with the foot of the perpendicular let fall from the vertex to the base. Begular Pyramids. 308 BOOK VII. SOLID GEOMETRY. 637. Dep. The perpendicular let fall from the vertex to the base of a regular pyramid is called the axis of the pyramid. The lateral edges of a regular pyramid are equal, for they cut off equal distances from the foot of the perpiendicular let fall from the vertex to the base. § 514 Therefore, the lateral faces of a regular pj'ramid are equal isosceles triangles. § 150 The altitudes of the lat- eral faces of a regular pyramid are ec[ual. 638. Def. The slant height of a regular pyramid is the alti- tude of any one of the lateral faces. It bisects the base of the lateral face in which it is dra\\Ti. § 149 639. Def. A truncated pjrramid is the portion of a pyramid included between the base and a section made bj- a plane cut- ting all the lateral edges. A frustum of a pyramid is the portion of a pyramid included between the base and a section parallel to the base. 640. Def. The al- titude of a frustum is the length of the per- pendicular between the planes of its bases. 641. Def. The lateral faces of a frustum of a regular pyra- mid are equal isosceles trapezoids ; and the sum of their areas is called the lateral area of the frustum. 642. Def. The slant height of the frustum of a regular pyramid is the altitude of one of these trapezoids. PYRAMIDS. 809 Proposition XIV. Theorem. 643. The lateral area of a regular pyramid is equal to half the product of its slant height by the perimeter of its base. Let S denote the lateral area of the regular pjrramid V-ABCDE, L its slant height, and P the perimeter of its base. To prove that S=^L X P. Proof. The A TAB, VBC, etc., are equal isosceles A. § 637 The area of each A is ^ i multiplied by its base. § 403 .•. the sum of the areas of these A is -^-Z X P. But the sum of the areas of these A is equal to 8, the lateral area of the pyramid. .'.S^\LXP. Q.E.D. 644, CoE. The lateral area of the frustum of a regular pyramid is equal to half the sum of the perimeters of the bases multiplied by the slant height of the frustum. § 407 Ex. 657. Find the lateral area of a regular pyramid if the slant height is 16 feet, and the base is a regular hexagon with side 12 feet. Ex. 658. Find the lateral area of a. regular pyramid if the slant height 1 is 8 feet, and the hase is a regular pentagon with side 5 feet. Ex. 659. Find the total surface of a regular pyramid if the slant height is.6 feet, and the base is a squaxe with side 4 feet. 310 BOOK VII. SOLID GEOMETRY. Proposition XV. Theorem. 645. If a pyrainid is cut hy a plane parallel to the base : 1. The edges and altitude are divided proportionally. 2. The section is a polygon similar to the base. Let V-ABCDE be cut by a plane parallel to its base, intersecting the lateral edges in a, b, c, d, e, and the altitude in o. 1. To prove that y2= f^' Vo ro' Proof. Since ahcde is II to ABODE, ab is II to AB, bo to BC ■■ -, and ao is II to AO. § 528 Va^ rb_ ^ Vo_ " VA'' VB ro §343 2. To prove the srrtion abode simUar to the base ABODE. Proof. Since A Vab is similar to A VAB, A Vbc is similar to A VBC, etc., § 354 ah _ f Vh\ _ be _( Vr. \ cd AB-\ Vb) ~'BC-\vc)=~CD' ^*°- ^ ^^^ .'. the homologous sides of the polygons are proportional. PYRAMIDS. 311 Since ah is II to AB, be to BC, cd to CD, etc., § 528 Aabe = Z. ABC, /Lhcd = Z. BCD, etc. § 534 Therefore, the two polygons are mutually equiangular. Hence, section ahcde is similar to the base ABCDE. § 351 Q.E.D. 646. CoK. 1. Any section of a pyramid parallel to the iase is to the base as the square of the distance from the vertex is to the square of the altitude of the pyramid. rO \VAJ AB yd" AB^ abode ah c ^-lo But = ^=-- §412 ABCDE AB" abode Vo Ax. 1 ABCDE VO 647. CoE. 2. If two pyramids having equal altitudes are cut hy planes parallel to the bases, and at equal distances from the vertices, the sections have the same ratio as the bases. ahcde Vo For = =5' ABCDE VO and = „ • § 646 A'B'C VO' But Vo = Vo', and VO = VO'. Hyp. .-. abode : ABCDE = a'b'c' : A'B'C. Whence ahcde : a'h'e' = ABCDE : A'B'C. § 330 648. CoE. 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to the 'bases, and at equal distances from the vertices, are equiva- 312 BOOK VII. SOLID GEOMETRY. Proposition XYI. Theoeem. 649. The volume of a triangular 'pyramid is the limit of the sum of the volumes of a series of inscribed or circumscribed prisms of equal altitude, if the number of prisms is indefinitely increased. Let S-ABC be a triangular pyramid. To prove that its volume is the limit of the sum, of the vol- umes of a series of inscribed or circumscribed prisms of equal altitude, if the number of prismas is indefinitely increased. Proof. Divide the altitude into n equal parts, and denote the length of each part by h. Through the points of division pass planes parallel to the base, cutting the pyramid in the sections DEF, GUI, etc. Through EF, HI, etc., pass planes parallel to SA, thus forming a series of prisms which have DEF, GUI, etc., for upper bases, and h for altitude. These prisms may be said to be inscribed in the pyramid. Again, through BC, EF, etc., pass planes parallel to SA, thus forming a series of prisms -which have ABC. DEF, etc., for lower bases, and h for altitude. These prisms may be said to be circumscribed a.bout the pyramid. PYRAMIDS. 313 Each inscribed prism is equal to the circumscribed prism directly above it. § 629 The difference, therefore, between the sum of the volumes of the inscribed prisms and the sum of the volumes of the circumscribed prisms is the prism D-ABC. Denote the volume of the pyramid by P, the. sum of tha volumes of the inscribed prisms by F, the sum of the volumes of the circumscribed prisms by V, and the difference between these two smns by d. Then V - V = d. By increasing n indefinitely, and consequently decreasing h indefinitely, d can be made less than any assigned quantity, however small, but cannot be made zero. Therefore, V — V can be made less than any assigned quan- tity, however small, but cannot be made zero. Now V > P, and rC eqvh'alpiif in a jiyramul, having for its alfitiirlp H. aii —^ = , > — = - ; and so on. v' O'B'' G'B'' V Vi «2 V2 vJ G'B'' § 671 Whence That is, V +Vi + V2 + ■ GB v' + vi' + v^' + --. v' G'B' V _ 'gb\ V'^lyB'^ §335 Q.B.D. 330 BOOK VII. SOLID GEOMETRY. REGULAR POLYHEDRONS. 673. Dbf. a regular polyhedron is a polyliedroii whose faces are equal regular polygons, and whose polyhedral angles are equal. Proposition XXIX. Pboblem. 674. To determine the number of regular convex poly- hedrons possible. A convex polyhedral angle must have at least three faces, and the sum of its face angles must be less than 360° (§ 581). 1. Since each angle of an equilateral triangle is 60°, convex polyhedral angles may be formed by combining three, four, or five equilateral triangles. The sum of six such angles is 360°, and hence greater than the sum of the face angles of a convex polyhedral angle. Hence, three regular convex polyhedrons are possible with equilateral triangles for faces. 2. Since each angle of a square is 90°, a convex polyhedral angle may be formed by combining three squares. The sum of four such angles is 360°, and therefore greater than the sum of the face angles of a convex polyhedral angle. Hence, one regular convex polyhedron is possible with squares. 3. Since each angle of a regular pentagon is 108° (§ 206), a convex polyhedral angle may be formed by combining three regular pentagons. The sum of four such angles is 432°, and therefore greater than the sum of the face angles of a convex polyhedral angle. Hence, one regular convex polyhedron is possible with regular pentagons. 4. The sum of three angles of a regular hexagon is 360°, of a regular heptagon is greater than 360°, etc. Hence, only five regular convex polyhedrons are possible. The five regular polyhedrons are called the tetrahedron, the hexahedron, the octahedron, the dodecahedron, the icosahedron. Q. E. F. REGULAR POLYHEDRONS. 331 675. The regular polyhedrons may be conistructed as follows : Draw the diagrams given below on stiff paper. Cut through the full lines and paste strips of paper on the edges shown in the diagrams. Fold on the dotted lines, and keep the edges in contact by the pasted strips of paper. 1 W fill II iL'^i ^' ^ /"' Tetrahedron. Hexahedron. Octahedron. Dodecahedron. Icosahedron. 332 BOOK VII. SOLID GEOMETRY. CYLINDERS. 676. DEr. A cylindrical surface is a curvefl surface generated by a straight line, wliich moves parallel to a fixed a straight line and con- stantly touches a fixed curve not in the plane of the straight line. The moving line is called the generatrix, and the fixed curve the directrix. 677. Def. The genera- trix in any position is Cylindrical Surface. called an element of the cylindrical surface. 678. Def. A cylinder is a solid bounded by a cjdindrical surface and two parallel plane surfaces. 679. Def. The two plane surfaces are called the bases, and the cylindrical surface is called the lateral surface. 680. Def. The altitude of a cylinder is the jierjiendicular distance between the planes of its bases. The elements of a cylinder are all equal. Kight Cylinder. ■' ^ 681. Def. A cylinder is a right cylin- der if its elements are perpendicular to its bases ; otherwise, an oblique cylinder. i *i 682. Def. A circular cylinder is a cyl- inder whose liases are cir'cles. 683. Def. A riglit circular cylinder is called a cylinder of revolution because oblique cylinder. CYLINDERS. 333 it may be generated by the revolution of a rectangle about one side as an axis. (ir^ ^''''- 684. Def. Similar cylinders of :, ||i|| revolution are cylinders generated I i|iy by the revolution of similar rect- lll I angles about homologous sides. ijlil 685. Def. a tangent line to a -''''''li cylinder is a straight line, not an ■ ■■'■ I . - 1 _ elument, which touches the lat- \ "1 r ^ff eral surface of the cylinder but > , ~ does not intersect it. 686. Def. a tangent plane to Inscribed Piism. g^ Cylinder is a plane which con- tains an element of the cylinder but does not cut the surface. The element contaiaed by the plane is called the element of contact. 687. Def. A prism is in- scribed in a cylinder when its lateral edges are elements of the cylinder and its leases are inscribed in the bases of the cylinder. 688. Def. A prism is cir- cumscribed about a cylinder when its lateral edges are paraUel to the elements of the cylinder jiEe===^ and its bases are circumscribed 1 , . -L 1 f? J 1 1 • n Circumscribed Prism. about the bases of the cylinder. 689. Def. A section of a cylinder is the figure formed hj its intersection with a plane passing through it. A right section of a cylinder is a section made by a pjlane per- pendicular to its elements. 334 BOOK VII. SOLID GEOMETRY. PEorosiTioN XXX. Th?:oeem. 690. Every section of a cylinder made hy a j^^ane passinej throurjh an element is a jxirallelogram. Let ABCD be a section of the cylinder AC made by a plane pass- ing through the element AD. To prove that ABCD is a parallelogram. Proof. Through B draw a line in the plane AC, II to AD. This line is an element of the cylindrical surface. § G76 Since this line is in both the plane and the cylindrical sur- face, it must lie their intersection and coincide with BC. Hence, BC coincides with a straight line parallel to AD. Therefore, BC is a straight line II to AD. Also AB is a straight line II to CD. § 528 .•. ABCD is a parallelogram. § 166 E.D. 691. Cdu. Evvrji ^eiiion of a rit/Jtt ci/Iiiii'lcr made by a ■plane pasxiiiij throuyh an element in a reetaiiyle. CYLINDERS. 335 Proposition XXXI. Thboebm. 692. The bases of a cylinder are equal. Let ABE and DCG be the bases of the cylinder AC. To prove that ABE = DCG. Proof. Let A, B, E be any three points in the perimeter of the lower base, and AD, BO, EG be elements of the surface. Draw AE, AB, EB, DG, DC, GO. Then AC, AG, EC are Z17. § 183 .-. AE = DG, AB = DC, and EB = GC. § 178 .-.A ABE = A DCG. §150 Place the lower base on the upper base so that the A ABE shall fall on the A DCG. Then A, B, E will fall on D, C, G. But A,B,E are awy points in the perimeter of the lower base. Therefore, all points in the perimeter of the lower base will fall on the perimeter of the upper base, and the bases will coincide and be equal. q. e.d. 693. CoE. 1. Any tivo parallel sections, cutting all the elements of a cylinder, are equal. For these sections are the bases of the included cylinder. 694. CoE. 2. Any section of a cylinder parallel to the base is equal to the base. 336 BOOK VII. SOLID GEOMETRY. PEOPosiTioif XXXII. Theorem. 695. If a prism whose hose is a regular polygon is imcrlhed in or vircumscnhed about a circular cylinder, and if the number of sides of the bctse of the prism is i?idefinitely increased, The volume of the cylinder is the limit of the volume of the prism. The lateral area of the cylinder is the limit of the lateral area of the prism. The perimeter of a right section of the cylinder is the limit of the perimeter of a right section of the prism. Let a prism whose base is a regular polygon be inscribed in a given circular cylinder, and a prism whose base is a regular poly- gon be circumscribed about the cylinder ; and let the number of sides of the base of the prism be indefinitely increased. To prove that the volume of the cylinder is the limit of the volume of the prism, that the lateral area of the cylinder is the limit of the lateral area of the prism, and that the perim- eter of a right section of the cylinder is the limit of the perimeter of a right section of the prism. CYLINDERS. 337 Proof. If the bases of the prism and cylinder could be made to coincide exactly, the prism and cylinder would coincide exactly ; and their volumes would be equal, their lateral areas would be equal, and the perimeters of their right sections would be equal. We cannot, however, make the bases of the prism and cylinder coincide exactly,, and we cannot, therefore, make the prism and cylinder coincide exactly ; but by increasing the number of sides of the base of the prism, we can make the base of the prism come as near coinciding with the base of the cylin- der as we choose (§ 464), and consequently make the prism come as near coinciding with the cylinder as we choose. Therefore, the difference between the volumes of the prism and the cylinder can be made less than any assigned value, however small, but cannot be made zero. The difference between the lateral areas of the prism and cylinder can be made less than any assigned value, however small, but cannot be made zero. The difference between the perimeters of the right sections of the prisin and cylinder can be made less than any assigned value, however small, but cannot be made zero. Therefore, the volume of the cylinder is the limit of the volume of the prism. § 275 The lateral area of the cylinder is the limit of the lateral area of the prism. § 275 The perimeter of a right section of the cylinder is the limit of the perimeter of a right section of the prism. § 275 Q. E. D. Note. This theorem can be proved true, when the base of the prLsm is not a regular polygon and the cylinder is not circular ; but it is not the province of Elementary Geometry to treat of cylinders whose bases are not circles. 338 BOOK VII. SOLID GEOMETRY. Pboposition XXXIII. Theorem. 696. The lateral area of a circular cylinder is equal to the product of the perimeter of a right section of the cylinder by an element. Let S denote the lateral area, P the perimeter of a right section, and E an element of the cylinder AC. To prove that S = P X E. Proof. Inscribe in the cylinder a prism with its base a regu- lar polygon, and denote its lateral area by ;S", and the perim- eter of its right section by P'. Then S' = P' X K § 607 If the number of lateral faces of the inscribed prism is indefinitely increased, S' approaches (S as a limit, § 695 and P' approaches P as a limit. § 695 .-. P' X E approaches P X ^ as a limit. § 279 But S' = P' X E, always. § 607 .-. S= P X E. §284 Q.E.D. 697. CoK. 1. The lateral area of a cylinder of revohdion is the product of the circumference of its base hy its altitude. CYLINDERS. 339 698. CoE. 2. If 8 denotes the lateral area, T the total area, H the altitude, and R the radius, of a cylinder of revolution, S=2'7rBxII; r = 27ri? X II + 2 irB^ = 2'7rIi(II + S). Peoposition- XXXIV. Theoebm. 699. The volume of a circular cylinder is equal to the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude, of the circular cylinder GA. To prove that V = B X H. Proof. Inscribe in the cylinder a prism with its base a regu- lar polygon, and denote its volume by F' and its base by B'. Then Y' = B' XH. § 628 If the number of its lateral faces is indelinitely increased, V approaches F as a limit, § 695 and B' approaches .B as a limit. § 454 .-. B' X JI approaches B X IT a.s a. limit. § 279 But r' = B' X H, always. § 628 .■.V=B X H. § 284 Q. E. D. 700. Cob. For a cylinder of revolution, with radius B, V=7rB^X H. 'AO BOOK VII. SOLID GEOMETRY. Proposition XXXV. Theorem. 701. The lateral areas, or the total areas, of similar cylinders of reoolution are to each other as the squares of their altittccles, or as the sc/uares of tJielr radii; and their volumes are to each other as the cubes of their altitudes, or as the cubes of their radii. Let S, S' denote the lateral areas, T, T' the total areas, V, V the volumes, H, H' the altitudes, R, R' the radii, of two similar cylin- ders of revolution. T„ prove that S : S' = T : T' = R- : IT'- = i?^ : R% in„l r : V = i?= : // '' = 7?" : It'^ Proof. Sinee the generating rectangles ai'e similar, we have by §§ 351, 335, if _ B _ R+B H'~B'~ II' + B'' Also we have b5r §§ G98, 700, 2'iTini B H B- ir- 2 irli'IV B' H' B'- R'"- T_ 'JirBdl+R) BfR+B\ B- R' T< 2 7rB'(R' + B') B'\R' + B'J B'^" R'^- r_ _ ttB-R _B^ R _ B^^ _ R^ V ~ irB'-R' ~B^-^r''~ B"" ^ IF'' Q.E.D. CYLINDERS. 341 PROBLEMS OP COMPUTATION. Ex. 678. The diameter of a well is 6 feet, aud the water is 7 feet deep. How many gallons of water are there in the well,'' reckoning 7i gallons to the cubic foot ? Ex. 679. When a hody is placed under water in a right ch-cular cylin- der 60 centimeters in diameter, the level of the water rises 40 centimeters. Find the volume of the body. Ex. 680. How many cubic yards of earth must be removed in construc1> ing a tunnel 100 yards long, whose section is a semicircle with a radius of 18 feet ? Ex. 681. How many square feet of sheet iron are required to make a fnnnp.1 18 inches iu diameter and 40 feet long ? Ex. 682. Find the radius of a cylindrical pail 14 inches high that will hold exactly 2 cubic feet. Ex. 683. The height of a cylindrical vessel that will hold 20 liters is equal to the diameter. Find the altitude and the radius. Ex. 684. If the total surface of a right circular cylinder is T, and the radius of the base is R, find the height of the cylinder. Ex. 685. If the lateral surface of a right circular cylinder is S, and the volume is V, find the radius of the base and the height. Ex. 686. If the circumference of the base of a right circular cylinder is C, and the height H, find the volume V. Ex. 687. Having given the total surface T of a right circular cylinder, m which the height is equal to the diameter of the base, find the vol- ume V. Ex. 688. If the circumference of the base of a right circular cylinder is C, and the total surface is T, find the volume V. Ex. 689. If the volume of a right circular cylinder is V, and the alti- tude is H, find the total surface T. Ex. 690. If F is the volume of a right circular cylmder in which the altitude equals the diameter, find the altitude H, and the total surface T. Ex. 691. If r is the total surface, and H the altitude of a right circular cylinder, find the radius B, and the volume V. 342 BOOK VII. SOLID GEOME'J'RY. CONES. 702. Def. a conical surface is the surface generated by a moving straight line which constantly touches a fixed curve and passes through a fixed point not in the plane of the curve. ^-~ - ''"• The moving straight line which gener- '~^'^^ ates the conical surface is called the gener- ^^ atrix, the fixed curve the directrix, and the H" fixed point the vertex. ^ ... 703. Def. The generatrix in any posi- ::'| ti(3n is called an element of the conical sur- ''% face. ^ If the generatrix is of indefinite length, i^S the surface consists of two portions, one -^^ a)ji ive and the other below the vertex, which "^^^ -msf are called the upper and lower nappes, respect- ivclv. CuniL-;il Surface. 704. Def. If the directrix is a closed curve, the solid bounded by the conical surface and a plane cutting all its elements is called a cone. The conical surface is called the lateral surface of the cone, and the plane surface is called the base of the cone. The vertex of the conical surface is called the vertex of the cone, and the elements of the conical sur'face are called the elements of the cone. The perpendicular distance *^°"'''- from the vertex to the plane of the base is called the altitude of the cone. CONES. 343 705, Def. a circular cone is a cone whose base is a The straight line joining tlie vertex and the centre base is called the axis of the cone. If the axis is perpendic- ular to the base, the cone is called a right cone. If the axis is oblique to the base, the cone is called an obUque cone. eoncs of Reyoh.tion. circle, of the 706. Def. a right circular cone is a cone '\\'hose base is a circle and whose axis is perpendicular to its base. A right circular cone is called a cone of revolution, because it may be generated by the revolution of a right triangle aljout one of its legs as an axis. The hypotenuse of the revolving triangle in any position is an element of the surface of the cone, and is called the slant height of the cone. The elements of a cone of revolution are all ec[ual. 707. Def. Similar cones of revolution are cones generated by the revolution of similar right trian- gles about homologous legs. 708. Def. A tangent line to a cone is a straight line, not an element, which touches the lateral surface of the cone but does not cut it. 709. Def. A tangent plane to a cone is a plane which contains an element of the cone but does not cut the surface. The element contained Ijy the plane Tangent Plane. is Called the element of contact. 3-14 BOOK VII. SOLID GEOMETRY. 710. Def. a pyramid is inscribed in a. cone when its lateral edges are elements of the cone and its base is inscribed in the base of the cone. 711. Def. A pyramid is circumscribed ^ about a cone when its base is circumscribed i'A about the base of the cone and its vertex ,,^* 1^ coincides with the vertex of the cone. , ,J' |^^ ' 712. Dee. A truncated cone is the por- ^ S^^ tion of a cone inchided between the base .^ ;; ^^^^^auS and a plane cutting all the elements. 'P -- ^^^^^^f A frustum of a cone is the portion of a ^ ^sm-' cone included between the base and a plane parallel to the base. iiis.rn...i pyramid. 713. Def. The base of the cone is called the lower base of the frustum, and the parallel section is called the upper base of the frustum. 714. Di:e. The altitude of a frustum of a cone is the perpendicular distance between the planes of its bases. ^■^--== 715. Def. The lateral surface of a frus- tum of a cone is the portion of the lateral surface of the cone included betn'een the bases of the frustum. Circumscribed I'yr.imiil. 716. Def. The elements of a cone between the bases of a frustimi of a cone of revolution are equal, and any one is called the slant height of the frustum. / A plane which cuts from the cone a C, frustum cuts from the inscribed or cir- cumscribed pyramid a frustum. frustum of a cone. CONES. 345 Pboposttiox XXXYI. Theorem. 717. Every section of a cone made hy a 'plcme pass- ing through its vertex is a triemgle. Let SBD be a section of the cone SBD made by a plane passing through the vertex S. To prove that SBD is a triatigle. Proof. BB is a straight line. § 506 Draw the straight lines SB and SB. These lines are elements of the surface of the cone. § 702 These lines lie in the cutting plane, since their extremities are in the plane. § 492 Hence, SB and SB are the intersections of the conical sur- face with the cutting plane. Therefore, the intersections of the conical surface and the plane are straight lines. Therefore, the section SBB is a triangle. § 117 Q.E.D. 346 BOOK VII. SOLID GEOMETRY. Proposition XXXVII. Theorem. 718. Every section of a circular cone made % a plane parallel to the base is a circle. Let the section abed of the circular cone S-ABCD be parallel to the base. To prove that alicil is a rh'de. Proof. Let lie tlie centre of the base, and let o be the point in which the axis SO pierces the plane of the parallel section. Through SO and any elements, SA, SB, pass planes cutting the base in the radii OA, OB, etc., aud the section abed in the straight lines oa, oh, etc. Then oa and oh are II, respectively, to OA and OB. § 528 Therefore, tlie A Soa. and Soh are similar-, respectively, to the A SOA and SOP,. § 354 • '"' f '^'"A "* c; o,-l ■■7i:r\-sf>) = oB- ^'"^ But OA = OB. § 217 .'. oa = ()/<, and ahed is a circle. § 216 QE.D. 719. (.'ok. Tlie a.rix of a dreiiJar cone passe.^ fhroui/Ii the centre of every section which is parallel to the base. CONES. 347 Pkoposition XXXVIII. Theoebm. 720. If a pyramid whose hase is a regular polygon is inscribed in or circumscribed about a circular cone, and if the number of sides of the base of the pyramid is indefinitely increased, the volume of the cone is the limit of the volume of the pyramid, and the lateral area of the cone is the limit of the lateral area of the pyramid. Let a pyramid whose base is a regular polygon be inscribed in a given circular cone, and a pyramid whose base is a regular polygon be circumscribed about the cone, and let the number of sides of the base of the pyramid be indefinitely increased. The proof is exactly the same as that of Prop. XXXII, if we substitute cone for cylinder and pyramids for prisms. 721. CoE. The volume of a frustum of a cone is the limit of the volumes of the frustums of the insorihed and circum- scribed pyramids, if the number of lateral faces is indefi- nitely increased, and the lateral area of the frustum of a cone is the limit of the lateral areas of the frustums of the inscribed and circumscribed pyramids. 848 BOOK VII. SOLID GEOMETRY. Peoposition- XXXIX. Theorem. 722. The lateral area of a cone of revolution is equal to half the 2)roduct of the slant height by the circumfer- ence of the base. Let S denote the lateral area, C the circumference of the base, and L the slant height, of the given cone. To prove that S = i C X L. Proof. Circumscribe about the cone a regular pyramid. Denote the perimeter of its base by P, and its lateral area hy S'. Then S' = iFxL. §643 If the number of the lateral faces of the circumscribed pyramid is indefinitely increased, S' approaches (S as a limit, and P approaches C as a limit. .'.^P X L approaches J C X i as a limit. But S' = i P X L. always. .■.S = iC XL. § 720 § 454 § 279 § 643 §284 Q.E.D.- 723. CoR. If S denotes the lateral area, T the total area, H the altitude, R the radius of the base, of a cone of revolu- **"'*' S = i{2TrExL)=7rBL; T =: ttBL + -ttB' ='7rIi(L + B). CONES. 349 Peopositiobt XL. Theobem. 724. The volume of a circular cone is equal to one third the product of its base hy its altitude. Let V denote the volume, B the base, and H the altitude of the given cone. To prove that V = ^ B % H. Proof. Inscribe in the cone a pyramid with a regular poly- gon for its base. Denote its volume by V, and its base by B'. Then r' = ^B'xH. §652 If the number of the lateral faces of the inscribed pyramid is indefinitely increased, V approaches F as a limit, § 720 and B' approaches i? as a limit. § 454 .-.^B' X H approaches ^5 X IT as a limit. § 279 But r' = iB' XH, always. § 652 .•.V=iBxir. §284 Q.E.D. 725. Cob. ^ R is the radius of the base, B = irRi § 463 .■.V = i'TrB^X H. 350 BOOK VII. S(_>LID GEOMETRY. Proposition XLI. Theorem. 726. The lateral areas, or the total areas, of tioo sim- ilar cones of revolution are to each other as the squares of their altitudes, as the squares of their radii, or as the squares of their slant heights ; and their volumes are to each other as the cubes of their altitudes, as the cubes of their radii, or as the cubes of their slant heights. Let S and S' denote the lateral areas, T and T' the total areas, V and V the volumes, H and H' the altitudes, R and R' the radii, L and L' the slant heights, of two similar cones of revolution. To prove that S : S' = T : T< = ff ^ . j-j(2 ^ jji . j^n^ and r:V' = R':II" = B^ : B" = L^:L'\ Proof H B L L+B IB B' L' L' + B' §§ 351, 335 ,9 itB.L B L B^ V- H^ ■wB'L' B' ^ V i?'2 ~ Z'2 ^ 7/'2' §723 T T'~ ivBd + 7?) B ,, L + B B- U ttBXL' + B') B' ^^ L< + B'~ B" ^ Z'^ IP ~ H'- r i7ri?^7f J?\^ II i?3 IP IJ i- irir-II' B'' " H'~ B'" ~ H'^ ~ i'3 § 725 Q. E. D. CONES. 351 Pkoposition XLII. Theoebm. 727. The lateral area of a frustum of a cone of revo- lution is equal to half the sum of the circumferences of its bases multiplied by the slant height. Let S denote the lateral area, C and c the circumferences of its bases, R and r their radii, and L the slant height. To prove that S = i(0 + d)X L. Proof. Circumscribe about the frustum of the cone a frus- tum of a regular pyramid. Denote the lateral area of this frustum by ;S", the perimeters of its lower and upper bases by P and p, respectively, and its slant height by L. Then S' = i(P+p)X L. § 644 If the number of lateral faces is indefinitely increased, S' approaches >S^ as a limit, § 721 and P + p approaches C + c as a limit. § § 454, 278 .'. i(P +p)L approaches ^(C + c)L as a limit. § 279 But S' = i(P+p)X L, always. § 644 .-. S = i(C' + c)X L. §284 Q.E.D. 728. CoE. The lateral area of a frustum of a cone of revo- lution is equal to the circumference of a section equidistant from its bases multiplied hy its slant height. 852 BOOK VII. SOLID GEOMETRY. Peoposition XLIII. Theorem. 729. The volume of a frustum of a circular cone is equivalent to the sum of the volumes of three cones ichose common altitude is the altitude of the frustum and whose bases are the lower base, the iqyper base, and the mean proportional between the bases of the frustum. Let V denote the volume, B the lower base, b the upper base, H the altitude of a frustum of a circular cone. To prove that V = i H{B + h + VBxb). Proof. Let V denote the volume, B' and b' the lower and upper bases, and If the altitude, of an inscribed frustum of a pyramid with regular polygons for its bases. Then F' = i -H"(-B' + 6' + V.B' X b'). § 667 If the number of the lateral faces of the inscribed frustum is indefinitely increased, V approaches 7' as a limit, § 721 B' approaches 5 as a limit, § 454 and b' approaches 5 as a limit. § 454 .-. B' X ¥ approaches B X b a& a, limit. § 281 ■■■ V^' X b' approaches V^ X b as a limit. § 283 .-. B' + b' + ■\/B' X b' approaches B + b + -s/B X 6 as a limit. § 278 CONES. 353 But r' = i H{B' + b' + V5' X V), always. § 667 .■.V = iH{B + h+^Bxb). §284 Q.E.D. 730. CoE. If the frustum is that of a cone of revolution, and R and r are the radii of its bases, B = ■irB\ b = Trr^. § 463 .-. VBxb = VTrWxTT? = irBr. .-. F = i 7r^(i?2 + 7^ + Br). Ex. 692. The radii of the bases of the frustum of a right circular cone are 20 Inches and 13 inches, respectively. If the altitude of the frustum is 15 inches and is bisected by a plane parallel to the bases, what is the lateral area of each frustum made by the plane ? Ex. 693. The radius of the base of a right circular cone is 8 feet, and the altitude is 10 feet. Find the area of its lateral surface, the area of its total surface, and the volume. Ex. 694. The height of a right circular cone is equal to the diameter of its base. Find the ratio of the area of the base to the area of the lat- eral surface. Ex. 695. The slant height of a right circular cone is 2 feet. At vfhat distance from the vertex must the slant height be cut by a plane parallel to the base, in order that the lateral surface may be divided into two equivalent parts ? Ex. 696. What does the volume F of a circular cone become, if the altitude is doubled ? If the radius of the base is doubled ? If both the altitude and the radius of the base are doubled ? Ex. 697. The slant height i of a right circular cone is equal to the diameter of the base. Find the total surface T. Ex. 698. If T is the total surface of a right circular cone whose slant height equals the diameter of the base, find the volume V. Ex. 699. If r is the total surface of a right circular cone, and R is the radius of the base, find the volume Y. Ex. 700. If T is the total surface of a right circular cone, and S is the lateral surface, find the volume V. 354 BOOK VII. SOLID GBOMETEY. THE PRISMATOID FORMULA. 731. Def. a polyhedron is called a prismatoid if it has for . bases two polygons in parallel planes, and for lateral faces triangles or trapezoids with one side common with one base and the opposite vertex or side common with the other base. 732. Def. The altitude of a prismatoid is the perpendicular distance between the planes of its bases. The mid-section of a prismatoid is the section made by a plane parallel to its bases and midway between them. The mid-section bisects the altitude and all the lateral edges. Proposition" XLIV. Theorem. 733. The volume of a prismatoid is equal to the prod- uct of one sixth of its altitude into the sum, of its bases and four times its mid-section. Let V denote the volume, B and b the bases, M the mid-section, and H the altitude, of a given prismatoid. To prove that V = i H{B + b + iM). THE PRISMATOID FORMULA. 355 Proof. If any lateral face is a trapezoid, divide it into two triangles by a diagonal. Take any point P in the mid-section and join P to the vertices of the polyhedron and of the mid-section. Divide the prismatoid into pyramids ■which have their vertices at P, and for their respective bases the lower base B, the upper base h, and the lateral faces of the prisma- toid. The lateral pyramid P-ADE is composed of three pyramids P-AFC, P-FCE, and P-FDE. Now P-AFC may be regarded as having vertex A and base PFG, and P-FCE, as having vertex E and base PEC. Hence, the volume of P-AFC is equal to I H X PFC, and the volume of P-FCE is equal to ^ -H" X PFC. § 661 The pyramid P-FDE is equivalent to twice P-FCE. For they have the same vertex P, and the base FDE is twice the base FCE, since the A FDE has its base DE twice the base FC of the A FCE (§ 406), and these triangles have the same altitude. Hence, the volume of P-FDE is equal to f -H" X PFC. Therefore, the volume of P-ADE, which is composed of P-AFC, P-FCE, and P-FDE, is equal to J S" X PFC. In like manner, the volume of each lateral pyramid is equal to f if X the area of that part of the mid-section which is included within it ; and, therefore, the total volume of all these lateral pyramids is equal to ^ H X M. The volume of the pyramid with base B is, I II Y. B, and the volume of the pyramid with base his\ H Xb. § 651 Therefore, V = ^ H (B -{- b -\- 4. M). q.e.d. 356 BOOK VII. SOLID GEOMETRY. 734. The prismatoid formula may he used for finding the volumes of all the solids of Elementary Geometry : The formula for the volume of the frustum of a pyramid is § 657 (1) v = i b:(£ + b + V^ X b). But ^II(B + b+-^BXb)=iII(B + b + 4:M). For if e and e' are corresponding sides of B and b, then i(e + e') is the corresponding side of M. Therefore, i(e + e') Addiag and reducing, Therefore, Squaring, and ■ ^/B 2_ 1~ -yfM 2 V^ = V5 + Vi. §190 §414 4 Jlf = 5 + & + 2 VfixT. If we put this value of 4 M in (1), the two members become identically equal. 735. If the base b becomes zero, we have a pyramid, and the prismatoid formula becomes V = \II Y. B. § 652 736. If the bases B and h are equal, we have a prism, and the prismatoid formula becomes V = H X B. § 628 Note. The Prismatoid Formula is takeR by permission from the work on Mensuration by Dr. George Bruce Halsted, Professor of Mathematics in the University of Texas. Ex. 701. Show that the prismatoid formula can be used for finding the volume of the frustum of a cone, for finding the volume of a cone, for finding the volume of a cylinder. EXERCISES. 357 FRUSTUMS OF PYRAMIDS AND OF CONES. Ex. 702. How many square feet of tin axe required to make a funnel, if the diameters of the top and bottom are 28 inches and 14 inches, respect- ively, and the height is 24 inches ? Ex. 703. rind the expense, at 60 cents a square foot, of polishing the curved surface of a marble column in the shape of the frustum of a right circular cone whose slant height is 12 feet, and the radii of whose bases are 3 feet 6 inches and 2 feet 4 inches, respectively. Ex. 704. The slant height of the frustum of a regular pyramid is 20 feet ; the sides of its square bases 40 feet and 16 feet. Find the volume. Ex. 705. If the bases of the frustima of a pyramid are regular hexagons whose sides are 1 foot and 2 feet, respectively, and the volume of the frus- tmn is 12 cubic feet, find the altitude. Ex. 706. The frustiun of a right circular cone 14 feet high has a volume of 924 cubic feet. Find the radii of its bases if their sum is 9 feet. Ex. 707. From a tight circular cone whose slant height is 30 feet, and the circumference of whose base is 10 feet, there is cut ofE by a plane parallel to the base a cone whose slant height is 6 feet. Find the lateral area and the volume of the frustum. Ex. 708. Find the difference between the volume of the frustum of a pyramid whose bases are squares, 8 feet and 6 feet, respectively, on a side and the volume of a prism of the same altitude whose base is a section of the frustiun parallel to its bases and equidistant from them. Ex. 709. A Dutch windmill in the shape of the frustum of a right cone is 12 meters high. The outer diameters at the bottom and the top are 16 meters and 12 meters, the inner diameters 12 meters and 10 meters. How many cubic meters of stone were required to build it ? Ex. 710. The chimney of a factory has the shape of a frustum of a regu- lar pyramid. Its height is 180 feet, and its upper and lower bases are squares whose sides are 10 feet and 16 feet, i;espectively. The flue is throughout a square whose side is 7 feet. How many cubic feet of material does the chimney contain? Ex. 711. Find the volume V of the fnistum of a cone of revolution, having given the slant height L, the height H, and the lateral area S. B68 BOOK vn. solid geometry. EQUIVALENT SOLIDS. Ex. 712. A cube eaok edge of which, is 12 inches is transformed into a right prism whose base is a rectangle 16 inches loug and 12 inches wide. Piud the height of the prism, and the difference between its total area and the total area of the cube. Ex. 713. The dimensions of a rectangular parallelopiped are a, 6, c. Pind (i) the height of an equivalent right circular oyUnder, having a for the radius of its base ; (ii) the height of an equivalent right ch-oular cone having a for the radius of its base. Ex. 714. A regular pyramid 12 feet high is transformed into a regular prism with an equivalent base. Eind the height of the prism. Ex. 715. The diameter of a cylinder is 14 feet, and its height is 8 feet. Find the height of an equivalent right prism, the base of which is a square with a side 4 feet long. Ex. 716. If one edge of a cube is a, what is the height B" of an equiva^ lent right circular cylinder whose radius is JJ ? Ex. 717. The heights of two equivalent right circular cylinders are in the ratio 4:9. If the diameter of the first is 6 feet, what is the diameter of the other ? Ex. 718. A right circular cylinder 6 feet in diameter is equivalent to a right circular cone 7 feet in diameter. If the height of the cone is 8 feet, what is the height of the cylinder ? Ex. 719. The frustum of a regular pyramid feet high has for bases squares 5 feet and 8 feet on a side. Eind the height of an equivalent regular pyramid whose base is a square 12 feet on a side. Ex. 720. The frustum of a cone of revolution is 5 feet high, and the diameters of its bases are 2 feet and 3 feet, respectively. Eind the height of an equivalent right circular cylinder whose base is equal in area to the section of the frustum made by a plane parallel to its bases and equidis- tant from the bases. Ex. 721. Eind the edge of a cube equivalent to a regular tetrahedron whose edge measures 3 inches. Ex. 722. Eind the edge of a cube equivalent to a regular octahedron whose edge measures 3 inches. EXERCISES. 359 SIMILAR SOLIDS. Ex. 723. The dimensions of a trunk are i feet, 3 feet, 2 feet. Find the dimensions of a trunk similar iu shape thai will hold four times as mucli. Ex. 724. By what number must the dimensions of a cylinder he multi- plied to obtain a similar cylinder (i) whose surface shall be ii, times that of the first ; (ii) whose volume shall be n times that of the first ? Ex. 725. A pyramid is cut by a plane parallel to the base which passes midway between the vertex and the plane of the base. Compare thq volumes of the entire pyramid and the pyramid cut off. Ex. 726. The height of a regular hexagonal pyramid is 36 feet, and one side of the base is 6 feet. What are the dimensions of a similar pyramid whose volume is ^jj that of the first ? Ex. 727. The length of one of the lateral edges of a pyramid is 4 meters. How far from the vertex will this edge be cut by a plane parallel to the base, which divides the pyramid into two equivalent parts ? Ex. 728. A lateral edge of a pyramid is a. At what distances from the vertex will this edge be out by two planes parallel to the base that divide the pyramid into three equivailent parts ? Ex. 729. A lateral edge of a pyramid is a. At what distance from the vertex will this edge be cut by a plane parallel to the base that divides the pyramid into two parts which are to each other as 3 : 4 ? Ex. 730. The volumes of two similar cones are 54 cubic feet and 432 cubic feet. The height of the first is 6 feet ; what is the height of the other ? Ex. 731. Two right circular cylinders have their diameters equal to their heights. Their volumes are as 3 : 4. Find the ratio of their heights. Ex. 732. Find the dimensions of a right circular cylinder if as large as a similar cylinder whose height is 20 feet, and diameter 10 feet. Ex. 733. The height of a cone of revolution is H, and the radius of its base is iJ. Find the dimensions of a similar cone three times as large. Ex. 734. The height of the frustum of a right cone is | the height of the entire cone. Compare the volumes of the frustum and the cone. Ex. 735. The frustum of a pyramid is 8 feet high, and two homologous edges of its bases are 4 feet and 3 feet, respectively. Compare the volume of the frustum and that of the entire pyramid. Book VIII. THE SPHERE. PLANE SECTIONS AND TANGENT PLANES. 737. Def. a sphere is a solid bounded by a surface all points of whicli are equally distant from a point witliin called the centre. 738. A sphere may be generated by the revolution of a semicir-cle A CB about its diameter AB as an axis. 739. Def. A radius of a sphere is a straight line di'awn from the centre to the surface. A diameter of a sphere is a straight line passing through the wntre and limited by the sm-face. 740. AH fill'' radii of a sphere are equal, and all the diame- ters nf a s/ihere are equal. 741. Def. A line or plane is tangent to a sphere when it has one, and oidy one, point in common wdth the surface of the sphere. 742. Def. Two spheres are tangent to each other when their surfaces have one, and only one, point in common. 360 THE SPHERE. 3G1 Proposition I. Theorem. 743. Every section of a sphere made by a plane is a circle. Let be the centre of the sphere, and ABD any section made by a plane. To prove that the sectioii ABD is a circle. Proof. Draw the radii OA, OB, to any two points A, B, in the boundary of the section, and draw O'C -L to the section. The rt. A OAC and OBC are equal. § 151 For 00 is common, and OA = OB. § 740 .■.CA=CB. §128 But A and B are any two points in tlie boundary of the sec- tion ; hence, all jjoints in the boundary are equally distant from C, and the section ABB is a circle. § 216 Q.E.D. 744. CoK. 1. The line joining the centre of a sphere to the centre of a circle of the sphere is perpendicular to the plane of the circle. 745. Cor. 2. Circles of a sphere made by planes equally distant from the centre are equal. Por Ig'' = AXf - 'OC'' ; and AO and 00 are the same for all equally distant circles ; therefore, AC is the same. 362 BOOK VIII. SOLID GEOMETRY. 746. CoE. 3. Of two circles made by planes unequally distant from the centre, the nearer is the greater. 747. Def. a great circle of a sphere is a section made by a plane which passes through the centre of the sphere. 748. Def. A small circle of a sphere is a section made by a plane which does not pass through the centre of the sphere. 749. Def. The axis of a circle of a sphere is the diameter of the sphere "which is perpendicular to the plane of the circle. The ends of the axis are called the poles of the circle. 750. CoE. 1. Parallel circles have the same axis and the same poles. 751. CoE. 2. All great circles of a sphere are equal. 752. CoE. 3. Every great circle bisects the sphere. For the two parts into which the sphere is divided can be so placed that they will coincide; otherwise there would be points on the surface unequally distant from the centre. 753. CoE. 4. Two great circles bisect each other. For the intersection of. their planes passes through the centre, and is, therefore, a diameter of each circle. 754. CoE. 5. If the planes of two great circles are perpen- dicular, each circle passes through the poles of the other. 755. CoK. 6. Through two given points on the surface of a sphere an arc of a great circle may always be drawn. For the two given points together with the centre of the sphere determine the plane of a great cii-cle which passes through the two given points. § 496 If, however, the two given points axe the ends of a diameter, the position of the circle is not determined. § 494 756. CoE. 7. Through three given points on the surface of a sphere one circle may be drawn, and only one. § 496 THE SPHERE. 363 757. Def. The distance between two points on the surface of a sphere is the arc of the great ciicle that joins them. Pkoposition II. Theoeem. 758. The distances of all j^oints in the circumference of a circle of a sjyhere from its poles are equal. 1111111 ilillll|||||f|illl|lli Let P, P' be the poles of the circle ABC, and A, B, C, any points in its circumference. To prove, that the, great circle arcs PA, PB, PC are equal. Proof. The straight lines PA, PB, PC are equal. § 614 Therefore, the arcs PA, PB, PC are equal. § 241 In like manner, the great circle arcs P'A, P'B, P'C may be proved equal. q.e.d. 759. Def. The distance on the surface of the sphere from the nearer pole of a small circle to any point in the circum- ference of the circle is called the polar distance of the circle. 760. Def. The distance on the surface of the sphere of a ffreat circle from either of Its poles is called the polar distance of the circle. 761. CoE. The polar distance of a great circle is a quad- rant ; that is, one fourth the circumference of a great circle. 56i BOOK VIII. SOLID GEOMETRY. PROPO.SITI0N III. ThE(JREM. 762. A point on the surface of a sphere, ivhich is at the distance of a quacbrmt from each of two other points, not the extremities of a diameter, is a pole of the great circde pjassing through these points. Let the distances PA and PB be quadrants, and let ABC be the great circle passing through A and B. l\i prooo tliiit P is a jwle of tit e great circle ABC. Proof. The .i POA and POB aie rt. A. § 288 .-. PO is _L to the pkuie of the O ABC. § 507 Hence, P is a poh; of the O ABC. § 749 Q.E.D. 763. ScHOLir:\r. The above tlieorem enables ns to cLeserilie with the compasses an are of a great cireL^ tlu'ougli two given points of tlie surface of a sphere. For, if with A and B as centres, and an opening of the compasses equal to the cliord of a cjuadrant of a great circle, we describe arcs, these arcs will intersect at a point P. Then, with P as centre, the arc pass- ing through A and B may be described. Til order to make the opening of the comjjasses eqrtal to the chord of a (piadrant of a great circle, the radius of the sphere must be known. THE SPHERE. 365 Proposition IV. Peoblbm. 764. Given a material sphere to find its diameter. ^c- \D- A.- ->^B- ">6 P'i' Let PBP'C represent a material sphere. It is required to find its diameter. From any point P of the given surface describe a circum- ference ABC on the surface. Then the straight line PB is known. Take three points A, B, and C in this circumference, and with the compasses measure the chords AB, BG, and CA. Construct the A A'B'C, with sides equal respectively to AB, EC, and CA, and circumscribe a O about the A. The radius D'B' of this O is equal to the radius of O ABC. m Construct the rt. A bdp, having the hypotenuse bp equal to BP, and one side bd equal to B'D'. Draw bp' _L to bp, laeeting pd produced in^'. Then^^' is equal to the diameter of the given sphere. Proof. Suppose the diameter PP'.and the straight line P'B drawn. The A BDP and bdp are equal. § 161 Hence, the A PBP' and pbp' are equal. § 142 Therefore, pp' = PP'. § 128 Q. E. F. 366 BOOK VIII. SOLID GEOMETRY. Proposition' V. Theorem. 765. A iilane perpeiicUcuIar to a radius at its ex- tremity is tangent tcj the sphere. Let be the centre of a sphere, and MN a plane perpendicular to the radius OP, at its extremity P. Til prnre tJiat MX is tiimjenf to the sphere. Proof. Let A lie any point except P in MN. Draw OA. Then OP < OA. § 512 Tlieref(ji-e, tlie point A is "without the sphere. § 737 But A is any point, except P, in the plane MX. .'. every point in MX, except P, is without the sphere. Therefore, MX is tangent to the sphere at P. § 741 Q.E.D. 766. Cor. 1. A pilanc tangent to a apliere is perpendicular to the radius drawn to tlie point of contact. 767. Cor. 2. ^1 line tangent to a circle of a sphere lies in the p>lane tangent to the spliere at the pioint of contact. § 508 768. Cor. 3. Aline in a tangent plane drawn tJirougJithe point of contact is tangent to the sphere at tliat point. 769. Cor. 4. The plane of two lines tangent to a sphere at the same point is tangent to the sphere at that point. THE SPHERE. 307 770. Det. a sphere is inscribed in a polyhedron when all the faces of the polyhedron are tangent to tlie sphere. 771. Def. a sphere is circumscribed about a polyhedron when all the vertices of the polyhedion lie in the siuface of the sphere. Peoposition VI. Theorem. 772. A sphere may he inscribed in any given tetra- hedron. Let A-BCD be the given tetrahedron. To prove that a sphere may he inscribed in A-BCD. Proof. Bisect the dihedral A at the edges AB, BI), and JB by the planes OAB, OBB, and OAD, respectively. Every point in the plane OAB is equally distant from the faces ABC and ABB. § 5.59 For a like reason, every point in the plane OBG is equally distant from the faces ABC and BBC; and every point in the plane OAC is equally distant from the faces ABC and ABC. Therefore, 0, the common intersection of these three planes, is equally distant from the four faces of the tetrahedron, and IS the centre of the sphere inscribed in the tetrahedron. § 770 O.E.D. 773, Cor. The six planes ivliieh Insect the sir dihedral angles of a tetrahedron intersect in the same point. 368 BOOK VIII. SOLID GEOMETRY. Proposition VII. Theorem. 774. A sphere may be circumscribed about any given tetrahedron. Let D-ABC be the given tetrahedron. To prove that a sphere may he circumscribed about D-ABC. Proof. Let M, X, respectively, be tlie centres of the circles circumscribed about the faces ABC, ACD. Let MB be ± to the face ABC, XS ± to the face ACB>. Then MB is the locus of points equidistant from A, B, C, and XS is the locus of points equidistant from A, C, I). § 516 Therefore, MB and XtS lie in the same plane, the plane _L to A C at its middle point. § 517 Also MB and XS, being _L to planes which are not II, can- not be II, and must therefore meet at some point 0. .". is equidistant from A, B, C, and D. Therefore, a spherical surface whose centre is 0, and radius OA, Avill pass through the points A, B, C, and I>. q.e.d. 775. Tor. 1. Tlie fotir perpendindars erected at the centres of the faces of a tetrahedron meet at the same point. 776. Cor. 2. The six planes perpendicniar to the edges of a tetrahedron at their middle points intersect at the same point. THE SPHERE. 369 Proposition- VIII. Theorem. 777. The intersection of two spherical surfaces is the circumference of a circle whose plane is perpendicular to the line joining the centres of the surfaces and whose centre is in that line. Let 0, C be the centres of the spherical surfaces, and let a plane passing through 0, 0' cut the spheres in great circles whose circum- ferences intersect in the points A and B. To prove that the spherical surfaces intersect in the circwm- ference of a circle whose plane is perpendicular to 00', and whose centre is the point C where AB meets 00'. Proof. The common chord AB is -L to 00' and bisected at C. § 264 If the plane of the two great circles is revolved about 00' as an axis, their circumferences ■will generate the two spherical surfaces, and the point A will describe the line of intersection of the surfaces. But during the revolution AC will remain constant in length and J. to 00'. Therefore, the line of intersection described by the point A will be the circumference of a circle whose centre is C and whose plane is J- to 00'. § 608 9.E.B, 370 BOOK VIII. SOLID GEOMETRY. FIGURES ON THE SURFACE OF A SPHERE. 778. Def. The angle of two curves passing through the same point is tlie angle formed by two tangents to the curves at that point. The angle formed by the intersection of two arcs of gi'eat circles of a sphere is called a spherical angle. Proposition IX. Theorem. 779. A sjyherical angle is measured by the arc of the great circle described from its vertex as a jyole and included between its sides {produced if necessary). Let AB, AC be arcs of great circles intersecting at A; AB' and AC, the tangents to these arcs at A ; BC the arc of the great circle described from A as a pole and included between AB and AC. Tfi prorii that tfie splierical Z. BAC is measured by arc BC. Proof. In the plane AOB, AB' is ± to ^0, and OB is ± to AO. .-. AB' is 11 to OB. Similarly, .4 C" is II to OC. .-. AB'AC = /.BOC. But A BOO is measured by arc BC. .' ■ Z^ B'AC" is measured hy arc BC. ■ ' . Z- BAC is measured by arc BC. §254 §288 § 104 § 534 § 288 Q.E.D. THE SPHERE. 371 780. Cob. A spherical angle has the same measure as the dihedral angle formed by the planes of the two circles. Peoposition X. Pkoblem. 781. To describe an arc of a great circle through a given point perpendicular to a given arc of a great circle. Let A be a point on the surface of a sphere, CHD an arc of a great circle, P its pole. Prom j4 as a pole describe an arc of a great circle cutting CUD at E. Prom ^ as a pole describe the arc AB through A. Then AB is the arc required. Proof. The axe AB is the arc of a great circle, and H is its pole by construction. § 762 The point JE is at the distance of a quadrant from P. § 761 Therefore, the arc AB produced ■will pass through P. Since the spherical Z. PBE is measured by the arc PE of a great circle (§ 779), the Z APE is a right angle. Therefore, the arc AB is J_ to the arc CHP. q.e.f. Ex. 736. Every point in a great circle wliioh bisects a given arc of a great circle at right angles is equidistant from the extremities of the given arc. 872 BOOK VIII. SOLID GEOMETRY. 782. Def. a spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of. great circles. The bounding arcs are the sides of the polygon ; the angles between the sides are the angles of the polygon ; the points of intersection of the sides are the vertices of the polygon. The values of the sides of a spherical polygon are usually expressed in degrees, minutes, and seconds. 783. The planes of the sides of a spherical polygon form a polyhedral angle whose vertex is the centre of the sphere, whose face angles are measured by the sides of the polygon, and whose dihedral angles have the same numerical measure as the angles of the polygon. Thus, the planes of the sides of the polygon ABCJD form the polyhedi'al angle 0-ABCD. The face angles AOB, BOC, etc., are meas- ured by the sides AB, BC, etc., of the polygon. The dihedral angle whose edge is OA has the same measure as the spherical angle BAD, etc. Hence, 784. From, any property of polyhe- dral angles we may infer an analogous property of spherical polygons ; and conversely. 785. Def. A spherical polygon is convex if the corresponding polyhedral angle is convex (§ 673). Every spherical poly- gon is assumed to be convex unless otherwise stated. 786. Def. A diagonal of a spherical polygon is an axe of a great circle connecting any two vertices which are not adjacent. 787. Def. A spherical triangle is a spherical polygon of three sides ; like a plane triangle, it may be right, obtuse, or acute ; equilateral, isosceles, or scalene. 788. Def. Two spherical polygons are equal if they can be applied, the one to the other, so as to coincide. THE SPHERE. 373 Proposition XI. Theoeem. 789. Each side of a spherical triangle is less than the sum of the other two sides. Let ABC be a spherical triangle, AB the longest side. To prove that . AB < AC + BC. Proof. In the corresponding trihedral angle 0-ABC, Z ^05 is less than Z^OC + Z^OC. §680 .■.AB 180° and < 540°. Proof. Since the A ABC, A'B'C, are polar A, A + a' = 180°, B + b' = 180°, C + <■' = 180°. § 793 .-. J + B + (' + a' + h' + <■< = 540°. Ax. 2 .-. A + /; + C = 510° - («' + /'' + '-•'). • Now a' + h' + <■' is less than 360°. § 790 .-. A + B + C = 510° — some value less than 300°. .-. A + B + C> 180°. Again a' + b' + c' is greater than 0°. .■.A + B+C<5i0°. Q.E.D. 796. CoE. A spherical triangle may have two, or even three, right angles; and it may have two, or even three, obtuse angles. 797. Def. a spherical triangle having two right angles is called a bi-rectangular triangle ; and a spherical triangle hav- ing three right angles is called a tri-rectangular triangle. 798. Def. The spherical excess of a triangle is the difference between the sum of its angles and 180°. THE SPHERE. 377 Proposition XVI. Thboebm. 799. In a hi-recta7igular spherical triangle the sides opposite the right angles are quadrants, and the side opposite the third angle measures that angle. 1 . " """"l Let PAB be a bi-rectangular triangle, with A, B right angles. To prove that PA and PB are quadrants, and that t]ieZ.P is measured hy the arc AB. Proof. Since the A A and B are right angles, the planes of the arcs PA, PB are _L to the plane of the arc AB. § 780 .'. PA and PB must each pass through the pole of AB. § 754 .'. P is the pole of AB, and PA, PB are quadrants. § 761 Also the Z P is measured by the arc AB. § 779 Q.E. D. 800. CoE. 1. If two sides of a .spherical triangle are quad- rants, the third side measures the opposite angle. 801. Cor. 2. Uach side of a tri-rectangular spherical tri- angle is a quadrant. 802. Cob. 3. Three pilanes passed through the centre of a sphere, each perjyendicular to the „v other two planes, divide the surface of the sphere into eight equal tri-rectangular triangles. 378 BOOK VIII. SOLID GEOMETRY. 803. Dep. If through the centre of a sphere three diam- eters AA', BB', CC are drawn, and the points A, B, C are joined by arcs of great circles, and also the points A', B', C, the two spherical triangles ABC and A'B'C are called symmetrical spherical triangles. In the same way we may form two symmetrical polygons of any number of sides, and place each of them in any position we choose upon the surface of the sphere. 804. Two symmetrical triangles are mutually equilateral and equiangular ; yet in general they cannot be made to coin- cide by superposition. If in the above figure the triangle ABC is made to slide on the surface of the sphere until the vertex A falls on A', it will be evident that the two triangles cannot be made to coincide and that the corresponding parts of the triangles occur in reverse order. 805. If, however, AB = AC, and A'B' = A'C'; that is, if the two symmetrical triangles are isosceles, then, because AB, AC, A'B', A'C are all equal, and the angles A and A' are equal, being opposite dihedral angles (§ 803), the two tri- angles can be made to coincide. There- B fore, 806. If two symmetrical spherical triangles are isosceles, they are superposable and therefore equal. THE SPHERE. 379 Peoposition XVII. Theorem. 807. Two symmetrical spherical triangles are equiv- alent. Let ABC, A'B'C be two symmetrical spherical triangles with their homologous vertices opposite each to each. To prove that the triangles ABC, A'B'C are equivalent. Proof. Let P be the pole of a small circle passing through the points A, B, C, and let POP' be a diameter. Draw the great circle arcs PA, PB, PC, P'A', P'B', PC. Then PA = PB = PC. §758 Now P'A< = PA, P'B' = PB, P'C = PG. § 804 .-. P'A' = P'B' = P'C. Ax. 1 .". the two symmetrical A PAC and P'A'C are isosceles. .■.APAC = A P'A'C. § 806 Similarly, A PAB = A P'A'B', and , APBC = AP'B'C'. Now AABC^APAC + APAB + APBC, As. 9 and A A'B'C ^ A P'A'C + A P'A'B' + A P'B'C. .-.AABC^AA'B'C. q.e.d. If the pole P should fall without the A ABC, then P' would fall without A A'B'C, and each triangle would be equivalent to the sum of two symmetrical isosceles triangles diminished by the third ; so that the result would be the same as before. 380 BOOK VIII. SOLID GEOMETRY. Proposition XVIII. Theokbm. 808. Tivo triangles on the same spliere or equal spheres are equal, if tioo sides and the included angle, or two angles and the included side, of the one are respectively equal to the corresp)onding parts of the other and arranged in the same order. Proof. By superposition, as in plane A. §§ 143, 1.39 Q.E. D. Proposition XIX. Theorem. 809. Two triangles on the same sphere or cqucd spheres are symonetricah if two sides and the included angle, or tvjo angles and. the included side, of the one are equcd, resi^ectively , to the corresjjonding parts of the other and arranged in the reverse order. Proof. Construct tlie A BEF' symmetrical witli respect to the A DEF upon tlie same sphere. THE SPHERE. 381 Then A ABC can be superposed upon the A DEF', so that they will coincide as in the corresponding case of plane A. But A BEF' and DEF are symmetrical by construction. .-. A ABC, which coincides with A DEF', is symmetrical with respect to A DEF. o e d Proposition XX. Theorem. 810. Two viutually equilateird triangles on tli.e same sphere or equal spheres are mutually equiangular, and are equal or symmetrical. Proof. The face A of the corresponding trihedral A at the centre of the sphere are equal respectively. § 2.37 Therefore, the corresponding dihedral A are equal. § 583 Hence, the A of the spherical A are respectively equal. Therefore, the A are equal or symmetrical, according as their equal sides are arranged in the same or reverse order. Q.E. D. Ex. 737. The radius of a sphere is 4 inches. From any point on the surface as a pole a circle is described upon the sphere with an opening of the compasses equal to 3 inches. Find the area of this circle. Ex. 738. The edge of a regular tetrahedron is a. Find the radii ii, R' of the inscribed and circumscribed spheres. Ex. 739. Find the diameter of the section of a sphere 10 inches In diameter made by a plane 3 inches from the centre. 382 BOOK VIII. SOLID GEOMETRY. Proposition XXI. Theoeem. 811. Tivo mutuaUy eqidangular triangles on the same sphere or equal sjjheres are mutually equilateral, and are either equal or symmetrical. Let the spherical triangles T and T' be mutually equiangular. To jirore that T and T' are mutually equilateral, and equal or si/i]> metrical. Proof. Let the A P be tlie polar A of T, and P' of T'. By hypothesis, the A T and T' are mutually equiangular. .•. the polar A F and J" are mutually equilateral. § 793 .'. the polar A F and F' are mutually equiangular-. § 810 But the A T and T' are the polar A of P and P'. § 792 .-. the A T and T' are mutually equilateral. § 793 Hence, the A T and T' are equal or symmetrical. § 810 Q.E.D. Note. The .<;tatement that mutually equiangular spherical triangles are mutually equilateral, and equal or .symmetrical, is true ouly when the>- are on the same sphere, or eiiual spheres. But when the spheres are unequal, the spherical triangles are imequal ; and the ratio of their homologous sides is equal to the ratio of the radii of the spheres. § 465 Ex. 740. At a given point in a given arc of a great circle, to construct a spherical angle equal to a given .spherical angle. Ex. 741. To inscribe a circle in a given spherical triangle. THE SPHERE. 383 Proposition XXII. Theorem. 812. In an isosceles S])herical triangle, the angles opposite the equal sides are equal. In the spherical triangle ABC, let AB equal AC. To prove that Z B = Z C. Proof. Draw the arc AB of a great circle, from the vertex A to the middle of the base BC. Then A ABB and ACB are inutuall}' ec|uilateral. .". A ABB and ACB are mutually equiangular. § 810 .■.ZB = ZC'. Q.E.D, 813. Cor. T7ie arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle of the base bisects the vertical angle, is perpendicular to the base, and divides the triangle into two symmetrical triangles. Ex. 742. To circunLscribe a circle about a giveu spherical triangle. Ex. 743. Given a spherical triangle whose sides are 60°, 80°, and 100°. Find the angles of its polar triangle. Ex. 744. Given a spherical triangle whose angles are 70°, 75°, and 95°. Find the sides of its polar triangle. Ex. 745. Find the ratio of two homologous sides of two mutually ecjui- angular triangles on spheres whose radii are 12 inches and 20 inches. 384 BOOK VIII. SOLID GEOMETRY. Proposition XXIII. Theorem. 814. If tivo angles of a sj^herical triangle are equal, the .sides opposite these angles are equal and the tri- angle is isosceles. In the spherical triangle ABC, let the angle B equal the angle C. Tn prnre tliat AC = AH. Proof. Let the A A'B'C be the pohu' A of the A ABC. Now ZII = Z C. Hyp. .-. A'C = A'B'. § 793 .-. ZB' = A C. § 812 .■:AC = AB. §793 Q.E. D. Ex. 746. To bisect a .spherical augle. Ex. 747. To construct a spherical triangle, having given two sides and the included angle. Ex. 748. To construct a spherical triangle, having given two augles and the included side. Ex. 749. To construct a spherical triangle, having given the three sides. Ex. 750. To construct a spherical triangle, having given the three angles. Ex. 751. To pass a plane tangent to a sphere at a given point on the surface of the sphere. Ex. 752. To pass a plane tangent to a .sphere through a given straight line withotrt the sphere. THE SPHERE. 385 Peoposition XXIV. Theoeem. 815. If tivo angles of a sjyJmical triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle ; conveesely, if two sides are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side. 1. In the triangle ABC, let the angle ABC be greater than the angle ACB. To prow that AC > AB. Proof. Draw the arc BI> of a great circle, making Z. CBD etjual Z ACB. Then DC = I)B. § 814 Now AD + DB > AB. § 789 .-. AD + DC > AB, or AC > AB. 2. Let AC be greater than AB. To prove that the Z. ABC is greater than the /- ACB. Proof. The A ABC must be equal to, less than, or greater than the A ACB. If Z ABC = Z C, then AC = AB; § 814 and if Z ABC is less than Z C, then AC < AB. (1) But both of these conclusions are contrary to the hypothesis. .■. Z ABC is greater than Z C. q. e.d. 386 BOOK VIII. SOLID GEOMETRY. Pkopositiok XXV. Theorem. 816. The shortest line that can he draion on the sur- face of a sphere between two points is the arc of a great circle, joining the two points not greater than a semi- circumference. Let AB be the arc of a great circle, not greater than a semicir- cumference, joining the points A and B. To prove that AB is the shortest line that can be drawn on the surf ace joining A and B. Proof. Let C be any point in AB. With A and B as poles and AC and BC as polar distances, describe two arcs DCF and ECG. The arcs DCF and FCG have only the point C in common. For if jF' is any other point in DCF, and if arcs of great circles AF and BF are drawn, then AF = AC. §758 But AF + BF>AC + BC. §789 Take away AF from the left member of the inequality, and its equal AC from the right member. Then BF>BC. Ax. 6 Therefore, BF> BG, the equal of BC. Hence, F lies without the circumference whose pole is B, and the arcs DCF and FCG have only the point C in common. THE SPHERE. 387 Kow let ADEB be any line from ^ to ^ on the surface of the sphere, which does not pass through C. This line will cut the arcs DCF and ECG in separate points D and E, and if we revolve the line AD about ^ as a fixed point until D coincides with C we shall have a line from A to G equal to the line AD. In like manner, we can draw a line from B to C equal to the line BE. Therefore, a line can be drawn from Ato B through C that is equal to the sum of the lines AD and BE, and hence less than the line ADEB by the line DE. Therefore, no line which does not pass through C can be the shortest line from A to B. Therefore, the shortest line from A to B passes through C. But C is awy point in the arc AB. Therefore, the shortest line from ^ to i> passes through every point of the arc AB, and consequently coincides with the arc AB. Therefore, the shortest line from ^ to 5 is the great circle arc^lB. Q.ED. Ex. 753. The three mediaxis of a spherical triangle meet in a point. Ex. 754. To construct a spherical surface with a given radius that passes through three given points. Ex. 755. To construct a spherical surface with a given radius that passes through two given points and is tangent to a given plane. Ex. 756. To construct a spherical surface with a given radius that passes through two given points and is tangent to a, given sphere. Ex. 757. All arcs of great circles drawn through a pole of a given great circle are perpendicular to the circumference of the great circle. Ex. 758. The smallest circle whose plane passes through a given point within a sphere is the one whose plane is perpendicular to the radius through the given point. BOOK VIII. SOLID GEOMKTRT. MEASUREMENT OF SPHERICAL SURFACES. 817. Def. a zone is a portion of the surface of a sphere included between two parallel planes. The circumferences of the sections niade by the planes are called the bases of the zone, and the distance between the planes is the altitude of the zone. 818. Def. A zone of one base is a zone one of whose bound- ing planes is tangent to the sphere. If a great circle FADQ (Fig. 1) is reTolved about its diam- eter PQ, the arc AD will generate a zone, the points A and D will generate its bases, and CF will be its altitude. The arc FA will generate a zone of one base. A 819. Def. A lune is a portion of the surface of a sphere bounded by two semicircumferences of great circles. 820. Def. The angle of a lune is the angle between the semicircumferences which form its boundaries. Thus (Fig. 2), ABECA is a lune, BAG is its angle. 821. Def. It is convenient to divide each of the eight equal tri-rectangular triangles of which the surface of a sphere is composed (§ 802) into 90 equal parts, and to call each of these parts a spherical degree. The surface of every sphere, there- fore, contains 720 spherical degrees. THE SPHERE. 389 Proposition XXVI. Theoebm. 822. The area of the surface generated hy a straight line revolving about an axis in its plane is equal to the product of the projection of the line on the axis hy the circwmference ivhose radius is a perpendicular erected at the middle point of the line and terminated hy the axis. A M B f\ \ ^\\ X c 6 D y Let XY be the axis, AB the revolving line, M its middle point, CD its projection on XY, MO perpendicular to XY, and MR to AB. To prove thai the area of surface by AB = CD x 2 irMR. Proof. 1. If AB is II to XY, CD = AB, MB coincides with MO, and the area AB is the surface of a right cylinder. § 697 2. If AB is not II to XY, and does not cut XY, the area AB is the surface of the frustum of a cone of revolution. -•. the area AB = AB X 2 irMO. § 728 Draw AE II to XY. The A ABE and MOB are similar. § 359 .■ . MO : AE = MR : AB. §361 .-. AB X MO = AE X MB. § 327 Or AB X MO = CD X MB. Substituting, the area AB = CD X 2 ttMB. 3. If A lies in the axis XY, the reasoning stiU holds, but AE and CD coincide, and the truth follows from § 722. Q.E.D. 390 BOOK VIII. SOLID GEOMETRY. Proposition XXVII. Theorem. 823. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let S denote the surface, R the radius, of a sphere generated by the semicircle ABCDE revolving about the diameter AE as an axis. To prove that S = AE X 2 ttB. Proof. Inscribe in the semicircle half of a regular polygon having an even number of sides, as ABODE. From the centre draw Js to the chords AB, BC, etc. These Js bisect the chords (§ 246) and are equal. § 249 Let a denote the length of each of these Js. From B, C, and D drop the Js BF, CO, and DG to AE. Then the area of surface by AB = AF x 2 tto., § 822 the area of surface by -BC = FO x 2 ira, etc. .-. the area of surface by ABCDE = AE x 2 ira. Denote the area of the surface described by the semi-polygon by S', and let the number of sides of the semi-polygon be indefinitely increased. Then S' approaches S as a, limit, and a approaches i? as a limit. .". AE X 2 TTO. approaches AE X 2 ttE as a limit. But S' = AE X 2 ira, always. .■.S = AEx 2irB. § 449 §279 §822 § 284 Q.E.D. THE SPHERE. 391 824. CoE. 1. The surface of a sphere is equivalent to four great circles ; that is, to 4 xB'^. For irli'^ is equal to the area of a great circle, § 463 and 4 tt^^ is equal to 2 ^ X 2 ttB, the area of the surface of a sphere. § 823 825. CoE. 2. The areas of the surfaces of two spheres are as the squares of their radii, or as the squares of their diameters. Let R and B' denote the radii, D and D' the diameters, and S and S' the areas of the surfaces of two spheres. Then S' 4 7ri?'2 R'-' {^D'f D'' 826. CoE. 3. The area of a zone is equal to the product of its altitude hy the circumference of a great circle. If we apply the reasoning of § 823 to the zone generated by the revolution of the arc BCD, we obtain the area of zone BCD = FG X lirR, where FG is the altitude of the zone and 2 irR the circumfer- ence of a great circle. 827. Cor. 4. Zones on the same sphere or equal spheres are to each other as their altitudes. 828. CoE. 6. A zone of one base is equivalent to a circle whose radius is the chord of the generating arc. The arc AB generates a zone of one base; and zone AB = AFx2-7rR = ttAF X AE. But AFXAF = AB\ § 370 .-. the zone AB = irAB . 392 BOOK VIII. SOLID GEOMETRY. Proposition XXVIII. Theokem. 829. The area of a lime is to the area of the surface of the sjAere as the number of degrees in the angle of the lune is to 360. Let ABEC be a lune, BCDF the great circle whose pole is A ; also let A denote the number of degrees in the angle of the lune, L the area of the lune, and S the area of the surface of the sphere. L:S = A : 360. To prove ttmt Proof. The ai-c BC measures the Z. A of the lune. § 779 Hence, arc DC : circuuifereiice BCDF = A : .360. If BC and BCDF are conmiensurable, let their common measiu'e be contained w times in BC, and ii times in BCDF. Then arc BC : circumference BCDF = m, : n. .-. A:?M) = w:n. ' §288 Pass arcs of great © througli the diameter AF and all the points of the division of BCDF. These arcs ^vill divide the entire surface into n, ecpial lunes, of which the lune ABFC will contain m. .' . L : S ■= III : 11. .-. L:S = .i :.3fiO. Ax. 1 If BC and BCT>F are incommensurable, the theorem can be joroved by the method of limits as in § 649. o e d THE SPHERE. 393 830. CoE. 1. The number of spherical degrees in a lune is equal to twice the number of aiigle degrees in the angle of the lune. If L and 8 are expressed in splierical degrees, § 821 then L:72Q = A: 360. § 829 Therefore, L = 2A. 831. CoE. 2. The area of a lune is equal to one ninetieth of the area of a great circle multiplied by the number of degrees in the angle of the lune: For L : 4 7r^2 ^ ^ . ggQ g ggg Therefore, L = 90 832. CoE. 3. Two lunes on the same sphere or equal spheres have the same ratio as their angles. For L:L' = : §831 90 90 ^ That is, L:L' = A: A'. 833. CoE. 4. Two lunes which have equal angles, but are situated on unequal spheres, have the same ratio as the squares of the radii of the spheres on which they are situated. That is, L:L' = R^: It'\ Ex. 759. Given the radius of a sphere 10 inches. Find the area of a lune whose angle is 30°. Ex. 760. Given the diameter of a sphere 16 inches. Find the area of a lune whose angle is 75°. 394 BOOK VIII. SOLID GEOMETRY. Peoposition XXIX. Theorem. 834. The area of a spherical triangle, expressed in spiherical degrees, is numerically equal to the spherical euiicess of the triangle. Let A, B, C denote the values of the angles of the spherical tri- angle ABC, and E the spherical excess. To ijrove, that the mnnhev of splierkal degrees in AABC = E. Proof. Produce the sides of the A ABC to complete circles. Tliese circles divide the surface of the sphere into eight si)herical triangles, of which any four having a common ver- tex, as A, form the surface of a hemisphere. The A A'BC, AB'C" are symmetrical and equivalent. § 807 And A ABC + A A'BC - lune ABA'C. Put the A AB'C for its equivalent, the A A'BC. Then A ABC + A AB'C =o= lune ABA'C. Also A ABC + A AB'C =o= lune BA B'C. And A ABC + A ABC ^ lune CA CB. Add and observe that in spherical degrees A ABC + A AB'C + A AB'C + A ABC = 360, § 821 and ABA'C + BAB'C + CA CB = 2 (A + B+ C). § 830 Then 2 A ABC + 360 = 2(A+ B+ C). .-. A ABC = A + B+C -180 = K q ^ j,. THE SPHERE. 395 835. CoK. 1. The area of a spherical triangle is to the area of the surface of the sphere as the number which expresses its spherical excess is to 720. For the number of spherical degrees in a spherical A ABC is equal to IS (§ 834), and the number of spherical degrees in S, the surface of the sphere, is equal to 720. § 821 .■.AABC:S=i::720. 836. Cob. 2. The area of a spherical triangle is equal to the area of a great circle multiplied by the number of degrees in E divided by one hundred eighty. For AABC:S=E:72Q. ' §835 But ^ = 4 itBP: § 824 4 irB'^E irR-'E •.AABC = 720 180 Ex. 761. What part of the surface of a sphere is a triangle whose angles are 120°, 100°, and 95° ? What is its area in square inches, if the radius of the sphere is 6 inches ? Ex. 762. Find the area of a spherical triangle whose angles are 100°, 120°, 140°, if the diameter of the sphere is 16 inches. Ex. 763. If the radii of two spheres are 6 inches and 4 inches respec- tively, and the distance between their centres is 5 inches, what is the area of the circle of intersection of these spheres ? Ex. 764. Find the radius of the circle determined in a sphere of 5 inches diameter by a plane 1 inch from the centre. Ex. 765. If the radii of two concentric spheres are B and R', and if a plane is drawn tangent to the interior sphere, what is the area of the section made in the other sphere ? Ex. 766. Two points A and B are 8 inches apart. Find the locus in space of a point 5 inches from A and 7 inches from B. Ex. 767. The radii of two parallel sections of the same sphere are a and 6 respectively, and the distance between these sections is d. Find the radius of the sphere. 396 BOOK VIII. SOLID GEOMETRY. Proposition XXX. Theorem. 837. If T denotes the nimiber vjJdch exj^resses the sum of the angles of a sjyhericcd polygon of n sides, the area of the 2)oh/gon expressed in spherical degrees is 7mmeri- calhj ecpialto T - (« - 2) 180. Let ABCDE be a polygon of n sides. To prove that the area of ABCDE exjiressed in spherical degrees, is numericalhj equal to T - {n - 2) 180. Proof. Divide the polygon into spherical triangles by draw- ing diagonals from any vertex, as A. These diagonals divide the polygon into j? — 2 spherical A. The area of each triangle in splierical degrees is luunerically equal to the sum of its angles minus 180. § 834 Hence, the sum of tlie areas of all the m — 2 triangles ex- jDressed in spherical degrees is nmuericalljr equal to the sum of all their angles minus (» — 2)180. Now the sum of the areas of the triangles is the area of the polygon, and the sum of the angles of the triangles is the sum i)f the angles of the polygon. Therefore, the area of the polygon expressed in spherical degrees is numerically equal to T — {i>, — 2) 180. q ^ £,. THE srilEUE. SPHERICAL VOLUMES. 838. Def. a spherical pyramid is the, pcjitioii of a spliere bounded by a spherical polygon and tliu planes of its sides. The centre of the sphere is the vertex of the pyramid, and the spherical poly- gon is the base of the pyramid. Thus, 0-ABC is a spherical pyramid. 839. Def. A spherical sector is the pfirtion of a si)here gen- erated liy the revolution of a circular sector about any diameter of the circle of whii'h the sector is a part. The base of a spherical sector is the zone generated liy the arc of the circular sector. 840. Def. A spherical segment is a portion of a sphere con- tained between two parallel planes. 841. Def. The bases of a spherical segment are the sec- tions made by the parallel planes, and the altitude of a spheri- cal segment is the perpendicular distam-e lictwei-n the bases. 842. Dkk. If one of the parallel planes is tangent to tlie spliere, the segment is called a segment of one base. 843. Def. A spherical wedge is a portion of a spliere bounded by a lune and two great semicircles. 598 B(.)()K VIII. SOLID GEOMETRY. Peoposition" XXXI. Thkokem. 844. The volume of a sjyJiere is equal to the produet of the area, of its surface by one third of its radius. Let R be the radius of a sphere whose centre is 0, S its surface, and V its volume. S X ili. Tn provr tliat V Proof. Conceive a cube to he circuinscribed about the sphere. Its vohime is greater than that of the S}ihere, because it con- tains the sphere. From 0, the centre of the sphere, conceive lines to be drawn to tlie vertices of the cube. These lines ai'e the eilges of six quadrangular pyramids, A\"liuse liases art- the fai^es of the cube, and whose t;ommon alti- tude is the radius of the sjihere. The volume of each pvramid is e(pral to .the product of its ] lase 1 1\' -J its altitude. Htnice, the volume of the six pyramids, that is, the volume of the circumscriVied cube, is equal to the area of the surface of the cube nudtiplied by -jfi?. Xow conceive planes drawn tangent to the sphere, at the points where tln' edges of the pyramids cut its surface. A^'e tlien have a circumscribed solid whose Arolnme is nearer that of the sphere than is the volume of the circumscribed culje, because each tangent plane cuts away a porti(_in of the cube. THE SPHERE. 399 From conceive lines to be drawn to each of the polyhe- dral angles of the solid thus formed. These lines form the edges of a series of pyramids, whose bases are together equal to the surface of the solid, and whose common altitude is the radius of the sphere ; and the Tolume of each pyramid thus formed is equal to the product of its base by -J its altitude. Hence, the sum of the volumes of these pyramids, that is, the volume of this new solid, is again equal to the area of its surface multiplied by -J R. If we denote the area of the surface of this polyhedron by jS", and the volume of the polyhedron by V, r'= S' XiA'. If we continue to draw tangent planes to the sphere, we con- tinue to diminish the circumscribed solid, since each new plane cuts off a corner of the polyhedron. By continuing this process indefinitely, we can make the difference between the volumes of the circumscribed solid and sphere less than any assigned quantity, however small, but we cannot make it zero ; and the difference between the areas of the surfaces of the circumscribed solid and sphere less than any assigned quantity, however small, but we cannot make it zero. Hence, F is the limit of V, and ^S is the limit of ,S". § 275 But r'= S' XiB, always. ■r=SxiB. §284 Q. ED. 845. CoE. 1. The volume of a sphere is equal to 4 ttS^ XiB; that !s, | ttS^, or ^ ttD^. (D = diameter.) 846. CoE. 2. The volumes of tivo spheres are to each other as the cubes of their radii. For V:r' = % irm : f Tri?'" = W' : B'\ 847. Cob. 3. The volume of a spherical piyramid is equal to the product of its base by one third of the radius of the sphere. 400 BOOK VIII. SOLID GEOMETRY 848. C(iR. 4. The vohiine of a si^heriral sector is equal to one third the product of the zone wltich forms its base hy the radius of the sjihere. If Jt denotes tlie i-;idiiis of tlie sjilierc, C tlie circumference of a great circle, 7/ tlie altitude of the zone, Z the surface of the zone, and V the voliinie of the sector ; then, (' = 2TrR (§ 458), Z = 2 ttR X H (§ 826), and V = 2 irRH x -^ i? = f ttR'H. Fkoposition XXXII. Problem. 849. To find the volume of a spherical segment. Let AC and BD be two semi-chords perpendicular to the diameter MN of the semicircle NCDM. Let OM be equal to R, AM to a, BM to b, AB to h, AC to r, BD to r'. Ca.sk 1. To find the ■volume in terms of Ti and a of the seq- ment of one luxe ijenerated In/ the cireular seni i-sei/nierif ACJI, OS the semicircle revolves alunif Jf.Y as <;«. axis. The sector generated liy OCJr= ^irRht. § 848 The cone generated hy OCA = ^irr- (R — a). § 724 Hence, segment AC 31 ■- - irir-o = l(2R^a i-irr^R-a) ~Rr"+ar-). K"o^r r" = AM X AN = a (2 R — a). .'. the segment ACM = ira" {R — - §370 (1) THE SPHERE. 401 If from the relation r^ = a(2R~a)we. find the value of R, and substitute it in (1), we obtain the volume in terms of the altitude and the radius of the base. The segment A CM = ^ -n-r'^a + ^ Tra\ (2) Case 2. To find the volume in terms of r, >•', and h of the segment of two bases generated by the circular semi-segment ABDC, as the semicircle revolves about N2I as an axis. Since the volume is obviously the difference of the volumes of the segments of one base generated by the circular semi- segments ACM and BDM, therefore, by formula (1), segment ABDC = vra^ f i? - ^ j - tt*^ fi^ - | j =^'n-R{a^-b^)-'^{a'-b^ (3) (put A for 0, - 5) = irRh (a + b)-^ (a" + ab + b^ = 7rA[(i?a + Rb)-l («= + ab + 6^]. Now a-b = h. .-. a^-2ab + b^ = hK Add 3 ab to each side, a^ + ab + b'' = h^ + 3 ab. Since (2R-a)a = r^ and (2R-b)b = r'^, § 370 ^2 + ,,'2 a' + b'' Ra + Rb=- 2 ' 2 ,-. the segment ^^X>C = ttA — ^ 1 ^ 3" ~ " p + r'2 , AV 7 '*' 7I = ttA 1—2— + ^ + ** ~ 3" ~ *^ J = - (-rrr' + irr") + —^ ■ 2^ ^6 (j.KF 402 BOOK VIII. SOLID GEOMETRY. NUMBEICAL EXERCISES. Ex. 768. Find the surface of a sphere, if the diameter is (1) 10 inches ; (ii) 1 foot 9 inches ; (iii) 2 feet 4 inclies ; (iv) 7 feet ; (v) 10.5 feet. Ex. 769. Find the diameter of a sphere if the surface is (i) 616 square inches ; (ii) 38i square feet ; (iii) iiy.JG square ftet. Ex. 770. The circumference of a dome, in the shape of a hemisphere is 66 feet. How many square feet of lead are required to cover it ? Ex. 771. If tlie ball on tlie top of St. Paul's Cathedral in London is 6 feet in diameter, ^Yhat would it cost to gild it at 7 cents per square inch ? Ex. 772. What is the numerical value of the radius of a sphere, if its surface has the same numerical value as the circumference of a great circle ? Ex. 773. Fmd the surface of a lune, if its angle is 30°, and the total surface of the sphere is 4 square feet. Ex. 774. What fractional part of the whole surface of a sphere is a spherical triangle whose angles are 43° 27', 81° 57', and 114° 36' ? Ex. 775. The angles of a spherical triangle are 60°, 70°, and 80°. The radius of the sphere is 14 feet. Find the area of the triangle. Ex. 776. The sides of a spherical triangle ai-e 80°, 74°, and 128°. The radius of the sphere is 14 feet. Find the area of the polar triangle in square feet. Ex. 777. Find the area of a spherical-polygon on a sphere whose radius is lOJ feet, if it.s angles are 10(1°. 120°, 140°, and 160°. Ex. 778. The planes of the faces of a quadrangular spherical pyramid make with each other angles of 80°, 100°, 120°, and 150° ; and the length of a lateral edge of the pyramid is 42 feet. Find the area of its base in square feet. Ex. 779. The planes of the faces of a. triangular spherical pyramid make with each other angles of 60°, 80°, and 100°, and the area of the base of the pyramid is 4 * square feet. Find the radius of the sphere. Ex. 7S0. The diameter of a sphere is 21 feet. Find the curved surface of a segment whose height is 5 feet. Ex. 781. In a sphere who.se radius is R, find the height of a zone whose area is equal to that of a great circle. NUMERICAL EXERCISES. 403 Ex. 782. What is the area of a zone of one base whose height is h, and the radius of tlie base r ? What would be the area if the height were twice as great ? Ex. 783. The altitude of the torrid zone is 3200 miles. Eind its area, assuming the earth to be a sphere with a radius of 4000 miles. Ex. 784. A plane divides the surface of a sphere of radius E into two zones, such that the surface of the greater is the mean proportional between the entire surface and the surface of the smaller. Eind the distance of the plane from the centre of the sphere. Ex. 785. If a sphere of radius E is cut by two parallel planes equally distant from the centre, so that the area of the zone comprised between the planes is equal to the sum of the areas of its bases, find the distance of either plane from the centre. Ex. 786. Eind the area of the zone generated by an arc of 30°, of which the radius is r, and which turns around a diameter passing through one of its extremities. Ex. 787. Eiad the area of the zone of a sphere of radius E, illuminated by a lamp placed at the distance h from the sphere. Ex. 788. How much of the earth's surface would a man see if he were raised to the height of the radius above it ? Ex. 789. To what height must a man be raised above the earth in order that he may see one sixth of its surface ? Ex. 790. The square on the diameter of a sphere and the square on an edge of the inscribed cube are as 3 : 1. Ex. 791. Eind the volume of a sphere, if the diameter is (i) 13 niches ; (ii) 3 feet 6 inches ; (ill) 10 feet 6 inches ; (iv) 14.7 feet. Ex. 792. Eind the diameter of a sphere, if the volume is (i) 75 cubic feet 1377 cubic inches ; (ii) 179 cubic feet 1162 cubic inches ; (iii) 1047.816 cubic feet ; (iv) 38.808 cubic yards. Ex. 793. Eind the volume of a sphere whose circumference is 45 feet. Ex. 794. Eind the volume T" of a sphere in terms of the circumference C of a great circle. Ex. 795. Eind the radius i? of a sphere, having given the volume V. Ex. 796. Eind the radius i? of a sphere, if its circumference and its volume have the same numerical value. 404 BOOK VIII. SOLID GEOMETRY. Ex. 797. The volume of a sphere is to the volume of the circumscribed ■ cube as * is to 6. Ex. 798. An iron ball 4 inches in diameter weighs 9 pounds. Find the weight of an ii-on shell 2 inches thick, whose external diameter is 20 inches. Ex. 799. The radius of a sphere is 7 feet. What is the volume of a wedge whose angle is 36° ? Ex. 800. What is the angle of a spherical wedge, if its volume is one cubic foot, and the volume of the entire sphere is 6 cubic feet ? Ex. 801. Find the volume of a spherical sector, if the area of the zone of its base is 3 square feet, and the radius of the sphere is 1 foot. Ex. 802. The radius of the base of a segment of a sphere is 16 inches, and the radius of the sphere is 20 inches. Find the volume of the segment. Ex. 803. The inside of a wash-basin is in the shape of the segment of a sphere ; the distance across the top is 16 inches, and its greatest depth is 6 inches. Find how many pints of water it wUl' hold, reckoning 71 gal- lons to the cubic foot. Ex. 804. What is the height of a zone, if its area is S, and the volume of the sphere to which it belongs is F ? Ex. 805. The radii of the bases of a spherical segment are 6 feet and 8 feet, and its height is 3 feet. Find its volume. Ex. 806. Find the volume of a triangular siAerical pyramid, if the angles of the spherical triangle which forms its base are each 100°, and the radius of the sphere is 7 feet. Ex. 807. The circumference of a sphere is 28 ff feet. Fuid the volume of that part of the sphere included by the faces of a trihedi-al angle at the centre, the dihedral angles of which are 80°, 105°, and 140°. Ex. 808. The planes of the faces of a quadrangular spherical pyramid make with each other angl&s of 80°, 100°, 1:20°, and 150°, and a lateral edge of the pyramid is P>1 feet. Find the volume of the pjrramid. Ex. 809. Having given the volume T", and the height h, of a spherical segment of one base, find the radius R of the sphere. Ex. 810. Find the weight of a sphere of radius E, which floats in a liquid of specific gravity s, with one fourth of its surface above the sur- face of the liquid. (The weight of a floating body is equal to the' weight of the liquid displaced.) MISCELLANEOUS EXERCISES. 405 MISCEILLANEOUS EXERCISES. Ex. 811. Determine a point in a given plane such that the difference of its distances from two given poiats on opposite sides of the plane shall he a maximum. Ex. 812. The portion of a tetrahedron cut ofE by a plane parallel to any face is a tetrahedron similar to the given tetrahedron. Ex. 813. Two symmetrical tetrahedrons are equivalent. Ex. 814. Two symmetrical polyhedrons may he decomposed into the same number of tetrahedrons symmetrical each to each. Ex. 815. Two symmetrical polyhedrons are equivalent. Ex. 816. If a solid has two planes of symmetry perpendicular to each other, the intersection of these planes is an axis of symmetry of the solid. Ex. 817. If a solid has three planes of symmetry perpendicular to one another, the three intersections of these planes are three axes of synmietry of the sohd ; and the common intersection of these axes is the centre of symmetry of the solid. Ex. 818. The volume of a sphere is to the volume of the inscribed cube as ;r is to jVs. Ex. 819. Find the area of the surface of the sphere inscribed in a regu- lar tetrahedron whose edge is 6 inches. Ex. 820. If a zone of one base is the mean proportional between the remainder of the surface of the sphere and the total surface of the sphere, find the distance of the base of the zone from the centre of the sphere. Ex. 821. Eind the difference between the volume of a frustum of a pyramid and the volume of a prism each 24 feet high, if the bases of the frustum are squares with sides 20 feet and 16 feet, respectively, and the base of the prism is the section of the frustum parallel to the bases and midway between them. Ex. 822. H the earth is assumed to be a sphere of 4000 miles radius, how far at sea can a lighthouse 100 feet high be seen ? Ex. 823. If the atmosphere extends 50 miles above the surface of the earth, and the earth is assumed to be a sphere of 4000 miles radius, find the volume of the atmosphere. 406 BOOK Vin. SOLID GEOMETRY. Ex. 824. Draw a line through the vertex of any trihedral angle, making equal angles with its edges. Ex. 825. In any trihedral angle, the three planes passed through the edges and the respective hisectors of the opposite face angles intersect in the same straight line. Ex. 826. In any trihedral angle, the three planes passed through the bisectors of the face angles, perpendicular to these faces, respectively, intersect in the same straight Une. Ex. 827. In any trihedral angle, the three planes passed through the edges, perpendicular to the opposite faces, respectively, intersect in the same straight line. Ex. 828. In a tetrahedron, the planes passed through the three lateral edges and the middle points of the opposite sides of the base intersect in a straight Hne. Ex. 829. The lines drawn from each vertex of a tetrahedron to the point of intersection of the medians of the opposite face all meet in a point called the centre of gravity, which divides each line so that the shorter seg- ment is to the whole line in the ratio 1 : 4. Ex. 830. The straight lines joining the middle points of the opposite edges of a tetrahedron all pass through the centre of gravity of the tetra^ hedron, and are bisected by the centre of gravity. Ex. 831. The plane which bisects a dihedral angle of a tetrahedron divides the opposite edge into segments proportional to the areas of the faces that include the dihedral angle. Ex. 832. The altitude of a regular tetrahedron is equal to the sum of the four perpendiculars let fall from any point within the tetrahedron upon the four faces. Ex. 833. Within a given tetrahedron, to find a point such that the planes passed through this point and the edges of the tetrahedron shall divide the tetrahedron into four equivalent tetrahedrons. Ex. 834. To cut a cube by a plane so that the section shall be a regular hexagon. Ex. 835. Two tetrahedrons are similar if a dihedral angle of one is equal to a dihedral angle of the other, and the faces that include these angles are respectively similar, and similarly placed. MISCELLANEOUS EXERCISES. 407 Ex. 836. To cut a tetrahedral angle so that the section shall be a parallelogram. Ex. 837. Two polyhedrons composed of the same nnniber of tetrahe- drons, similar each to each and similarly placed, are similar. Ex. 838. If the homologous faces of two similax pyramids are respect- ively parallel, the straight lines which join the homologous vertices of the pyramids meet in a point. Ex. 839. The volume of a right circidar cylinder is equal to the product of the lateral area by half the radius. Ex. 840. The volume of a right circular cylinder is equal to the product of the area of the rectangle which generates it, by the length of the cir- cumference generated by the point of intersection of the diagonals of the rectangle. Ex. 841. If the jiltitude of a right circular cylinder is equal to the diameter of the base, the volume is equal to the total area multiplied by a third of the radius. Ex. 842. Show that the prismatoid formula can be used for finding the Tolmne of a sphere. Ex. 843. Find the altitude of a zone equivalent to a great circle. Ex. 844. Find the area of a spherical pentagon whose angles are 122°, 128°, 131°, 160°, 161°, if the sui-face of the sphere is 150 square feet. Construct a spherical surface with given radius : Ex. 845. Passing through a, given point and tangent to two given Ex. 846. Passing through a given point and tangent to two given spheres. Ex. 847. Passing through a given point and tangent to a given plane and a given sphere. Ea 848. Tangent to three given planes. Ex. 849. Tangent to three given spheres. Ex. 850. Tangent to two given planes and a given sphere. Ex. 851. Tangent to two given spheres and a given plane. Ex. 852: Through a given point to pass a plane tangent to a given cir- cular cylinder. 408 BOOK VIII. SOLID GEOMETRY. Ex. 853. Through a given point to pass a plane tangent to a given circular cone. Ex. 854. Find the radius and the surface of a sphere whose volume is one cubic yard. Ex. 855. Pind the centre of a sphere -whose surface passes through three given points, and touches a given plane. Ex. 856. Find the centre of a sphere whose surface touches two given planes, and passes through two given points which lie between the planes. Ex. 857. The volume of a sphere is two thirds the volume of the cir- cumscribed circular cylinder, and its surface is two thirds the total surface of the oyliuder. Ex. 858. Given a sphere, a cylinder circumscribed about the sphere, and a cone of two nappes inscribed in the cylinder. If any two planes are drawn perpendicular to the axis of the three figures, the spherical segment between the planes is equivalent to the difference between the corresponding cylindrical and conic segments. Ex. 859. A sphere 12 inches in diameter has -an auger hole 3 inches in diameter through its centre. Find the remaining volume. Ex. 860. Find the area of a solid generated by an equilateral triangle turning about one of its sides, if the length of the side is a. Ex. 861. Compare the volumes of the solids generated by a rectangle turning successively about two adjacent sides,>.the lengths of these sides being a and 6. Ex. 862. An equilateral triangle revolves about one of its altitudes. Compare the lateral area of the cone generated by the triangle and the surface of the sphere generated by the inscribed circle. Ex. 863. An equilateral triangle revolves about one of its altitudes. Compare the volumes of the solids generatedby the triangle, the inscribed circle, and the circumscribed circle. Ex. 864. The perpendicular let fall from the point of intersection of the medians of a given triangle upon any plane not cutting the triangle is equal to one third the sum of the perpendiculars from the vertices of the triangle upon the same plane. Ex. 865. The perpendicular from the centre of gravity of a tetrahedron to a plane not cutting the tetrahedron is equal to one fourth the sum of the perpendiculars from the vertices of the tetrahedron to the plane. Book IX. CONIC SECTIONS. THE PARABOLA. 850. Def. a parabola is a curve which is the locus of a point that moves in a plane so that its distance from a fixed point in the plane is always equal to its distance from a fixed line in the plane. 851. Def. The fixed point is called the focus ; and the fixed line, the directrix. 852. A parabola may be described by the continuous motion of a point, as follows : A F Place a ruler so that one of its edges shall coincide with the directrix DK Then place a right triangle with its base edge in contact with the edge of the ruler. Fasten one end of a string, whose length is equal to the other edge BC, to the point B, and the other end to a pin fixed at the focus F. Then slide the triangle BCJE along the directrix, keeping the string tightly pressed against the ruler by the point of a pencil P. The point P will describe a parabola ; for during the motion we always have PF equal to PC. 409 410 BOOK IX. CONIC SECTIONS. Proposition I. Pkoblem. 853. To construct a parabola hy points, having given its focus and its directrix. c . .^^ o L //'' i / i D / IM a[. H l\. E \<^ Q '^ Let F be the focus, and CDE the directrix. Draw FD _L to CU, meeting CU at D. Bisect FD at J. Then ^ is a point of the curve. § 850 Through, any point M in the line DF, to the right of A, draw a line II to CK With F as centre and DM as radius, draw arcs cutting this line at the points F and Q. Then P and Q are points of the curve. Proof. Draw FC and QF ± to CF. Then FC = DM, and QE = DM, § 180 and DM = FF = QF. Const. .-. FC = FF, and QF = QF. Ax. 1 Therefore, F and Q are points of the curve. § 860 In this way any number of points may be found ; and a con- tinuous curve drawn through the points thus determined is the parabola whose focus is F and directrix CDF. q.e.f. THE PARABOLA. 411 854. Def. The point A is called the vertex of the curve. The line DI" produced indefinitely in both directions is called the axis of the curve. 855. Def. The line FF, joining the focus to any point F on the curve, is called the focal radius of F. 856. Def. The distance AM is called the abscissa, and the distance FM the ordinate, of the point F. 857. Def. The double ordinate LS, through the focus, is called the latus rectum or parameter. 858. CoE. 1. The parabola is symmetrical with respect to its axis. § 210 Tor FF = FQ (Const.), and, therefore, FM = QM. § 149 859. CoH. 2. The curve lies entirely on one side of the perpendicular to the axis erected at the vertex; namely, on the same side as the focus. For any point on the other side of this perpendicular is obviously nearer to the directrix than to the focus. 860. CoK. 3. The parabola is not a closed curve. Eor any point on the axis of the curve to the right of F is evidently nearer to the focus than to the directrix. Hence, the parabola QAF cannot cross the axis to the right of F. 861. CoE. 4. The latus rectum is equal to 4,AF. § 850 For, if iff is drawn ± to CDF, then LF = LG, and LG = DF. .■.LF= DF = 2AF. Similarly, BF = DF = 2AF. Therefore, LR = 4.AF. Note. In the following propositions, the focus will he denoted hy F, the vertex hy A, and the point where the axis meets the directrix by D. 412 BOOK IX. CONIC SECTIONS. Proposition II. Theoeem. 862. The ordinate of any point of a parabola is the mean proportional between the latus rectum and the abscissa. § 860 Let P be any point, AM its abscissa, PM its ordinate. To prova that FM^ = 4 AF X AM. Proof. PM'' = FF'^ - FM^ = I)M^ - FM^" = (DM - FM) (DM + FM) = DF (DF + FM + FM) = 2AF(2AF + 2 FM) = 2AF(2AM). Hence, FM^ = iAF X AM. Q.E.D. 863. CoE. 1. The greater the abscissa of a point, the greater the ordinate. 864. CoE. 2. The squares of any ttvo ordinates are as the abscissas. For 1^ = "^^FxAM _ AM _ QN' 4 AF X AN AN THE PARABOLA. 418 Pkoposition III. Theorem. 865. Every point within the parabola is nearer to the focus than to the directrix; and every point without the parabola is farther from the focus than from the directrix. a §138 D A F 1. Let Q be a point within the parabola. Draw QC perpendicu- lar to the. directrix, cutting the curve at P. Draw QF and PF. To prove that QF < QC. Proof. In the A QFF, QF Q'C. Proof. In the A Q'FP, Q'F > PF - PQ'. .■.Q'F>PC -PQ'>Q'C. §138 Q.E.D. 866. Cob. A point is within or without a parabola accord- ing as its distance from the focus is less than, or greater than, its distance from the directrix. 867. Dee. A straight line which touches, but does not cut, a parabola, is called a tangent to the parabola. The point where it touches the parabola is called the point of contact. 414 BOOK IX. CONIC SECTIONS. Proposition IV. Theobbm. 868. If a line PT is drawn from any point P of the curve, bisecting the angle between PF and the perpen- dicular from P to the directrix, every point of the line PT, except P, is without the airve. ^9 1 ^ c I j 5^ G xy 1 E jPiw r \ 1 1 Ty y r V V A p i 1 . N Let PC be the perpendicular from P to the directrix, the angle FPT equal the angle CPT, and let X be any other point in PT except P. To prove that X is without the curve. Proof. Draw XE ± to the directrix, and dra-w CX, FX, CF. Let CF meet PT at P. In the isos. A P CF, CR = PF, and CX = FX. But FX 2 a. Proof. Let P be any point on the curve between QF and QF. Draw PF and PF'. Then QF + QF' > PF + PF'. § 100 But PF + PF' = 2 a. § 896 Therefore, QF + QF > 2 a. 2. Let Q' be a point within the curve. To prove that Q'F.+ Q'F <2 a. Proof. Let P be any point of the curve included between FQ' produced and F'Q' produced. Then Q'F + Q'F' < PF + PF'. § 100 That is, Q'F + Q'F' <2a. q.e.d. 922. Cor. Conversely, a point is without or within an ellipse according as the sum of its distances from the foci is greater than or less than 2 a. 923. Dee. A straight line which touches but does not cut an ellipse is called a tangent to the ellipse. The point where a tangent touches the ellipse is called the point of contact. 434 BOOK IX. CONIC SECTIONS. Peoposition XVIII. Theokem. 924. If through a point P of an ellipse a line is drawn bisecting the angle between one of the focal radii and the other produced, every point in this line except P is with- out the curve. T A' F' O F Let PT bisect the angle FTG between F'P and FP produced, and let Q be any point in PT except P. To prove thdbt Q is without the curve. Proof. Upon FP produced take PG equal to PF'. Draw GF, QF, QF', QG. Then QG+QF>GF. §138 Now AGPQ = A F'PQ. § 143 Therefore, QG = QF'. § 128 Also GF=2a. Const. Therefore, QF' +QF>2a. Therefore, Q is without the curve. § 922 Q.E.D. 925. Cor. 1. The bisector of the angle between one of the focal radii from any point P and the other produced through P is a tangent to the curve at P. r 923 926. Cor. 2. The tangent to an ellipse at any point bisects the angle between one focal radius and the other produced. THE ELLIPSE. 435 927. CoE. 3. If GF' euts FT at X, then OX = F'X, and FT is perpendicular to GF'. § 161 928. Cob. 4. The locus of the foot of a perpendicular from the focus of an ellipse to a tangent is the auxiliary circle. For F'X = GX, and F'0 = OF. Therefore, OX = \FG = \(2a) = a. § 189 Therefore, the point X lies in the auxUiary circle. Pkoposition XIX. Problem. 929. To draw a tangent to an ellipse from an external point. , ^ Let the arcs drawn with P as centre and PF as radius, and with F' as centre and 2 a as radius intersect in G and S. Draw GF' and SF', cutting the curve in Q and R, respectiYely. Draw QP and BP, and they will be the tangents required. Proof. By construction, PG = PF, and QG = QF. § 896 .-. A PQG = A PQF. § 150 .-.APQG^APQF. §128 Therefore, PQ is the tangent at Q. § 926 For like reason PR is the tangent at R. q.e.f. 930. CoE. Two tangents may always he drawn to an ellipse ffeom an external point. 436 BOOK IX. CONIC SECTIONS. Proposition XX. Theorem. 931. The tangents drawn at two corresponding points of an ellipse and its auxiliary circle cut the major axis ■produced at the same point. y^ G "^^^A 2 /^ ^ ^X \ .--^ ^ '\ 1 ^^*»^ ND M F Let the tangent to the auxiliary circle at Q cut the major axis produced at T, and let the ordinate QM of the circle meet the ellipse at P. Draw TG through P. To prove that TG is the tangent to the ellipse at P. Proof. Through S, any point in the ellipse except P, draw SD ± to AA' ; and let DS produced cut TG in R, the auxil- iary circle in K, and the tangent at Q in L. Then But Again, But : PM = DT : MT = LD : QM, §§ 356, 351 RD : LD = PM : QM. §330 PM:QM=h:a. §917 .-. RD : LD = b:a. Ax. 1 SD:KD = h: a. §917 .-. RD:LD=SD: KB. Ax. 1 LD > KD. Ax. 8 .-. RD > SD. .'. jB is -without the ellipse. ence, PT is the tangent at P. §923 Q.E.D. THE ELLIPSE. 437 932. CoE. 1. 0TX0M= a". § 367 933. Dbf. The straight liae FN drawn through the point of contact of a tangent, perpendicular to the tangent, is called the normal. 934. Dei-. MT is called the subtangent, MIf the subnormal. 935. CoE. 2. The normal bisects the angle between the focal radii of the point of contact. For Z TPN- = Z GPJST = 90°. § 933 But Z TPF = Z GPF'. § 926 .-. Z FPN = Z F'PN^. Ax. 3 Hence, a ray of light issuing from F will be reflected to F'. 936. CoK. 3. If d denotes the abscissa of the point of con- tact, the distances measured on the major axis from the centre c? to the tangent and the normal are -r and e'^d, respectively. § 932 § 367 Since OM = d, and OT X OM = a\ therefore, a Since 0MXMT = QM\ and MN XMT = PM\ therefore. OM QM' a^ MN PM' V Therefore, OM - MN a^-V c' _ 2 OM ■ a^ a' ^ That is. ON 0M~^- Hence, 0N=e^y< OM = e^d. § 333 438 BOOK IX. CONIC SECTIONS. Proposition XXI. Theorem. 937. The tangents drawn at the ends of any diameter are parallel to each other. ' Let POQ be any diameter, PT and QT' the tangents at P, Q, respec- tively, meeting the major axis at T, T' To prove that PT is II to QT'. Proof. Draw the ordinates PM, QN. Then A 0PM = A OQN (§ 141), and OM = ON. § 128 a^ But OT = = ^';^,andOT'-^^(§936). ■.0T = OT' Therefore, A OPT = A OQT', §143 and Z OPT = Z OQT'. §128 Hence, PT is II to QT'. - - §111 Q.E.D. 938. Def. One diameter is conjugate to another, if the fii'st is parallel to the tangents at the extremities of the second. Thus, is. BOS is II to PT, RS is conjugate to PQ. THE ELLIPSE. 439 Proposition^ XXII. Theorem. 939. If one diameter is conjugate to a second, the second is conjugate to the first. Let the diameter POP' be parallel to the tangent RT'. To prove that ROR' is parallel to the tangent PT. Proof. Draw the ordinates PM and RN, and produce them to meet the auxiliary circle in Q and S, respectively. Draw OP, OQ,OS; and draw the tangents QT, ST'. Now, since OP is II to RT', the A OMP and T'JfR are simUar. § 354 .-. T'JSf: OM=NR: MP. § 351 But NR:NS = MP : MQ, § 917 or NR:MP=NS .MQ. §330 .-. T'N: OM = NS : MQ. Ax. 1 Hence, A T'NS and OMQ are similar. § 357 .■.ZNT'S = ZMOQ. §351 .-. T'S is II to OQ. § 114 Hence, AQOS = Z 08T' = 90°. § 264 .-. SO is II to QT. § 104 440 BOOK IX. CONIC SECTIONS. . A SNO and QMT are similar. §354 .■ . ON : TM = NS : MQ, §351 = NB : MP. § 917 . A ONB and TMP are similar. §357 .-. OB is 11 to FT. §114 .'. BB' is conjugate to PI". §938 Q.E.D, 940. CoK. 1. Angle QOS is a right angle. 941. Cob. 2. MP : ON = b : a. For OS = OQ, §217 and since ZNST' = Z.MQO, § 176 ZNSO = ZMOQ. §84 Hence, ANSO = AMOQ. §141 .■.ON = MQ. § 128 .-. MP : ON = MP: MQ. But MP ■.MQ = b:a. §917 Hence, MP ■.ON=b:a. Ax. 1 THE ELLIPSE. 441 Proposition- XXIII. Theorem. 942. The area of an ellipse is equal to -n-ab. A. M N let A'PRA be any semi-ellipse. To prove thai the area of twice A'PRA is equal to iraib. Proof. Let FM, RN be two ordinates of the ellipse, and let Q, S be the corresponding points on the auxiliary circle. Draw PV, QU W to the major axis, meeting US in V, U. Then the area of O PN = PM X MN, § 398 and the area of O Qlf = QM X MN. Z\ § 398 £7 PN _ PM X MN _ PM ~ QM' Therefore, §917 O QN QM X MN i^M a The same relation will be true for all the rectangles that can be similarly drawn iu the ellipse and auxiliary circle. -rr sum of ZI7 in ellipse h Hence, j-=-t r^ = - ■ § 335 sum or lU m circle a And this is true whatever be the number of the rectangles. But the limit of the sum of the UJ in the ellipse is the area of the ellipse, and the limit of those in the is the area of the O. area of ellipse _ h Therefore, area of circle §285 Therefore, the area of the ellipse = - X ira^ = irab. § 463 * Q.E.D. 442 BOOK IX. CONIC SECTIONS. Proposition XXIV. Theorem. 943. The section of a right circular cone made hy a plane that cuts all the elements of the surface of the cone is an ellijose. Let APA' be the curve traced on the surface of the cone SBC by a plane that cuts all the elements of the surface of the cone. To prove that the curve APA' is an ellipse. Proof. The plane passed tlirough. the axis of the cone _L to the secant plane APA' cuts the surface of the cone in the ele- ments SB, SC, and the secant plane in the line AA'. Describe the © and 0' tangent to SB, SC, AA'. Let the points of contact be i), IJ, F, and B, C, F', respectively. Tui-n BSC and the © 0, 0' about the axis of the cone. The lines SB, SC will generate the surface of a right circular cone cut by the secant plane in the curve APA' ; and the © 0, 0' will generate spheres which touch the cone in the © DNH, BN'G, and the secant plane in the points F, F'- THE ELLIPSE. 443 Let P be any point on the curve APA'. Draw PF, PF' • and draw SP, which touches the © DH, 5C at the points N, IP, respectively. Since PF and PiV are tangent to the sphere O, they are tangent to the circle of the sphere made by a plane passing through P, F, and iV. Therefore, PF = PJSf. § 261 Likewise, PF' = FN' § 261 Hence, PF + PF' = PiV + PN' = NN', a constant quantity. § 716 Therefore, APA' is an ellipse with the points F and F' for foci, and AA' as 2 a. q.e.d. 944. Cor. If the secant plane is parallel to the base, the section is a circle. Ex. 885. The major axis is the longest chord that can be drawn in an ellipse. Ex. 886. If the angle FBT is a right angle, then a^ = 2 6". Ex. 887. To draw a tangent and a normal at a given point of an ellipse. Ex. 888. To draw a tangent to an ellipse parallel to a given straight line. Ex. 889. Given the foci ; to describe an ellipse touching a given straight line. Ex. 890. Prove that OF^ = OT x ON. (See figure, page 436.) Ex. 891. Prove that OM : ON = a^ : A (See figjire, page 436.) Ex. 892. The minor axis is the shortest diameter of an ellipse. Ex. 893. At what points of an ellipse will the normal pass through the centre of the ellipse ? Ex. 894. If FR, F'S are the perpendiculars dropped from the foci to any tangent, then FB x F'S = 6^. 444 BOOK IX. CONIC SECTIONS. Ex. 895. The semi-mmor axis of an ellipse is the mean proportional between, the segments of the major axis made by one of the foci. Ex. 896. The area of an ellipse is to the area of its auxiliary circle as the minor axis is to the major axis. Ex. 897. To draw a diameter conjugate to a given diameter in a given ellipse. Ex. 898. Given 2 a, 2 6, one focus, and one point of an ellipse, to con- struct the ellipse. Ex. 899. If from a pomt P a pair of tangents PQ and PB are drawn to an ellipse, then PQ and PB subtend equal angles at either focus. Ex. 900. If a quadrilateral is circumscribed about an ellipse, either pair of its opposite sides subtend angles at either focus whose sum is equal to two right angles. Ex. 901. To find the foci of an ellipse, having given the major axis and one point on the curve. Ex. 902. To find the foci of an ellipse, having given the major axis and a straight line which touches the curve. Ex. 903. If a straight line moves so that its extremities are always in contact with two fixed straight lines perpendicular to each other, then any point of the moving line describes an ellipse. Ex. 904. To construct an ellipse, having given one of the foci and three tangents. Ex. 905. To construct an ellipse, having given one focus, two tangents, and one of the points of contact. Ex. 906. To construct an ellipse, having given one focus, one vertex, and one tangent. Ex. 907. The area of the parallelogram formed by the tangents to an ellipse at the extremities of any pair of conjugate diameters is equal to the area of the rectangle contained by the axes of the ellipse. Ex. 908. Given an ellipse, to find by construction the centre, the foci, and the axes. Ex. 909. The circle described on any focal radius of an ellipse as a diameter is tangent to the auxiliary circle. Ex. 910. If the ordinate and the tangent at any point P of an ellipse meet a diameter at H and K, respectively, then OH x OK = OQ'^. Q is -the point in which the diameter cuts the curve. THE HYPERBOLA. 445 THE HYPERBOLA. 945. Def. An hyperbola is a curve ■which is the locus of a point that moves in a plane so that the difference of its dis- tances from two fixed points in the plane is constant. 946. Def. The fixed points are called the foci, and the straight lines which join a point of the locus to the foci are called the focal radii of that point. 947. The constant difference of the focal radii is denoted by 2 a, and the distance between the foci by 2 c. 948. Def. The ratio c:a is called the eccentricity, and is denoted by e. Therefore, c = ae. 949. Cob. 2 a must be less than 2 c (§ 138) ; hence, e must be greater than 1. 950. An hyperbola may be described by the continuous motion of a point, as follows : To one of the foci F' fasten one end of a rigid bar F'B so that it is capable of turning freely about F' as a centre in the plane of the paper. 446 BOOK IX. CONIC SECTIONS. Take a string whose length is less than that of the bar by the constant difference 2 a, and fasten one end of it at the other focus F, and the other end at the extremity B of the bar. If no-w the rod is made to revolve about F' while the string is kept constantly stretched by the point of a pencil at P, in contact with the bar, the point P will trace an hyperbola. For, as the bar revolves, F'P and FP are each increasing by the same amount ; namely, the length of that portion of the string which is removed from the bar between any two positions of P ; hence, the difference between F'P and FP will remain constantly the same. The curve obtained by turning the bar about F' is the right- hand Irannh of the hyperbola. Another similar branch on the left may be described in the same manner by making the bar revolve about i'' as a centre. If the two branches of the hyperbola cut the line FF' at A and A\ then, from the symmetry of the construction, AA} — 2 a. The hyperbola, therefore, consists of two similar branches which are separated at their nearest points by the distance 2 a, and which recede indefinitely from the line FF' and from one another. THE HYPERBOLA. 447 Proposition XXV. Problem. 951. To construct an hyperbola by points, having given the fod and the constant difference 2 a. Let F, F' be the foci, and CD equal a. Lay off OA equal to OA' equal to CD. Then A and A' are two points of the curve. Proof. Prom the construction, AA' = 2a and AF = A'F'. Therefore, AF' - AF = AF' - A'F' = AA' = 2 a. And A'F - A'F' = A'F ~ AF =AA' = 2a. To locate other points, mark any point X in F'F produced. Describe arcs with F' and F as centres, and A'X and AX as radii, intersecting in P, Q. Then P and Q are points of the curve. By describing the same arcs with the foci interchanged, two more points B and S may be found. By assuming other points in F'F produced, any number of points may be foimd ; and the curve passing through all these points is an hyperbola having F, F' for foci and 2 a for the constant difference of the focal radii. q.e.f. 448 BOOK IX. CONIC SECTIONS. 952. Cob. 1. iVb point of the curve can he situated on the perpendicular to FF' erected at 0. For every point of this _L is equidistant from the foci. 953. Def. The point is called the centre ; AA' is called the transverse axis ; A and A' are called the vertices. 954. Def. In the J- to FF' erected at 0, let B, B' be two points at a distance from A (or A') equal to c ; then BB' is called the conjugate axis, and is denoted by 2 b. 955. Def. If the transverse and conjugate axes are equal, the hyperbola is said to be equilateral or rectangular. 956. CoE. 2. Both the axes are bisected at the centre. 957. CoE. 3. By § 371, c^ = a"" + b\ 958. CoE. 4. The curve is symmetrical with respect to the transverse axis. 959. Def. The distances of a point of the curve from the transverse axis and the conjugate axis are called respectively the ordinate and abscissa of the point. The double ordinate through the focus is caUed the latus rectum or parameter. Note. The letters A, A', B, B', F, F', and O will be used to designate the same points as in the above figure. THE HYPERBOLA. 449 Proposition XXVI. Theorem. 960. An hyperbola is symmetrical loith respect to its corijugate axis. B / F' A' o A F Let P be a point of the curve, PDQ be perpendicular to OB, meet- ing OB at D, and let DQ equal DP. To prove that Q is also a point of the curve. Proof. Join P and Q to the foci F, F'^ Turn ODQF' about OD ; F' will fall on F, and Q on P. Therefore, QF' = PF, and Z PQF' = Z QPF. Therefore, A PQF' = A QPF, §143 and QF = PF'. §128 Hence, QF - QF' = PF' - PF. Ax. 3 But PF' -PF=^2a. Hyp. Therefore, QF -QF' = 2 a. Ax. 1 Therefore, Q is a point of the curve. §945 Q.E.D. 961. Dep. Every chord passing through the centre is called a diameter. 962. Cor. 1. An hyperhola consists of four equal quad- rantal arcs symmetrically placed about the centre. § 213 963. CoK. 2. Uvery diameter is bisected at the centre. § 209 450 BOOK IX. CONIC SECTIONS. Proposition XXVII. Theorem. 964. If d denotes the abscissa of a point of an hyper- bola, r and r' its focal radii, then r = ed — a, and r' = ed + a. F' a: A. F Let P be any point of the hyperbola, PM perpendicular to AA', d equal OM, r equal PF, r' equal PF'- To prove that r = ed — a, r' = ed -{- a. Proof. Prom the rt. A FPM, F'PM, r' = PM' + FM\ §371 : PM + FM . 72 Therefore, r'^ - r^ = F'M' - FM". Or (r' + r) (r' - r) = {F'M + FM) {F'M - FM). Now r' — r = 2 a., and F'M — FM = 2 c. Also F'M + FM = 2 OF + 2 FM ^20M=2d. By substitution, a(r' + r) = 2 ed. 2cd a From r' + r — 2ed, and /•' — r = 2 a, by addition, 2r' = 2(6d + a); by subtraction, 2r=^2{ed — a). Therefore, r = ed — a,andr' = ed + a. q.e.d. Or J,' _l_ J. . ■ = 2ed. THE HYPERBOLA. 451 965. Def. The circle described upon AA' as a diameter is called the auxiliary circle. Proposition XXVIII. Theorem. 966. Any ordinate of an hyperbola is to the tangent from its foot to the auxiliary circle as b is to a. Let P be any point of the hyperbola, PM the ordinate, MQ the tangent drawn from M to the auxiliary circle. To prove that PM: QM = h:a. Proof. Let OM equal d. Then QM'' = (P- a?. § 372 Also PM" = PF' - FM" = {6d-af-(d-cY §964 = eW -2aed + a'-d^ + 2cd- c\ Or since e = ae, and a^-c^^-h'', §§ 948, 957 PM ={&'-!) d^'-h^' Therefore, ¥m'' : QM^ = V- : a?. Or PM : QM =b -.a. ,(d^-a^. Q. E. D. 452 BOOK IX. CONIC SECTIONS. Pkoposition XXIX. Theorem. 967. The square of the ordinate of a point in an hyperhola is to the product of the distances from the foot of the ordinate to the vertices as b^ is to d. Let P be any point of the hyperbola, PM the ordinate, MQ the tangent from M to the auxiliary circle. To prove that PM^ : AM X A'M = b^ : a\ Proof. Now Tm'' : 'QM'' = W : a\ § 966 But 'QM'' = AM X A'M. § 381 Therefore, Tm'' : AM XA'M=b^: a\ q. e. d. 968. CoK. The lotus rectum is the third proportional to the transverse and conjugate axes. For LF : AF X A'F = b^ : a^. But AF =c — a, and AF' = c -\- a. Therefore, AF X A'F = c^ — a^ =. b^ Hence, Z^' ib^^b^: a\ And LF:b =b -.a. Therefore, 2a:2b = 2b:2LF. § 967 §957 THE HYPERBOLA. 453 Proposition XXX. Theorem. 969. The difference of the distances of any point from the foci of an hyp.erhola is greater than or less than 2 a, according as the point is on the concave or convex side of the curve. F' A' A. F 1. Let Q be a point on the concave side of the curve. To prove that QF' —QF>2a. Proof. Let QF' meet the curve at P. F'Q = F'P + PQ (Ax. 9), and FQ F'P - FP. Ax. 6 But F'P-FP = 2a. §947 Therefore, F'Q- FQ>2a. 2. Let Q' be a point on the convex side of the curve. To prove that Q'F' - Q'F <2a. Proof. Let Q'F cut the curve at P. F'Q' < F'P + Pgi (§ 138), and FQ' = FP + PQ'. Ax. 9 .-. F'Q' - FQ' < F'P - FP. Ax. 5 But F'P - FP =2a. § 947 Therefore, F'Q' - FQ' <2a. q, e. d. 970. CoK. Conversely, a point is on the concave or the con- vex side of the hyperbola according as the difference of its distances from the foci is greater than or less than 2 a. 454 BOOK IX. CONIC SECTIONS. 971. Dbf. a straight line -wMcli touches but does not cut the hyperbola is called a tangent, and the point where it touches the hyperbola is called the point of contact to the hyperbola. Proposition XXXI. Theorem. 972. If through a point P of an hyperbola a line is drawn bisecting the angle between the focal radii, every point in this line except P is on the convex side of the curve. F' A' Let PT bisect the angle FPF', and let Q be any point in FT except P. To prove that Q is on the convex side of the curve. Proof. Take PG equal to PF; draw FG, QF, QF', QG. Then QF'-QG< GF'- Also APGQ = APFQ(IU3);.-.QG^QF. Also GF' = PF'-PF = 2 a. Therefore, QF' - QF < 2 a. Therefore, Q is on the convex side of the curve. §338 §970 Q. E. D. 973. Cor. 1. The bisector of the angle between the focal radii from any point P is the tangent to the curve at P. § 971 974. Cor. 2. The tangent to an hyperbola at any point bisects the angle between the focal radii drawn to that point. 975. CoK. 3. The tangent at A is perpendicular to AA'. THE HYPERBOLA. 455 976. CoE. 4. If FG outs FT at X, then GX = FX, and FT is perpendicular to FG. § 161 977. CoE. 6. The locus of the foot of the perpendicular from the focus to a tangent is the auxiliary circle. For FX = GX, and FO = OF- .-. OX=iF'G = i(PF'-FF)=a. §189 Therefore, tlie point X lies on the auxiliary circle. Peopositioit XXXII. Peoblem. 978. To draw a tangent to an hyperbola from a given point P on the convex side of the hyperbola. Let the arcs described -with P as centre and PF as radius, and with F' as centre and 2 a as radius intersect in G and B. Draw F'G and F'R, and produce them to meet the curve in Q and D, respectively. Draw PQ and PD. PQ and PD are the tangents required. Proof. PG = PF,^F=QF' -2a=QG. .■.APQG = APQF. §150 .■.ZPQG = ZPQF. §128 .•. PQ is the tangent at Q. § 973 Per like reason, PD is the tangent at D. q.e.p. 456 BOOK IX. CONIC SECTIONS. 979. Cor. Two tangents may he drawn to an hyperbola from a point on the convex side of the hyperbola. 980. Def. If a rectangle is constructed with its adjacent sides equal, respectively, to the transverse and conjugate axes of the hyperbola, and with one side tangent to the curve at A and its opposite side at A', its diagonals produced are called the asymptotes of the hyperbola. Proposition XXXIII. Theorem. 981. The asymptotes of an hyperbola never meet the curve, however far produced. Let TR be an asymptote of the hyperbola whose centre is 0. To prove that TR never meets the curve. Proof. Let G be the intersection of arcs described from and F' as centres with OF and 2 a, respectively, as radii. THE HYPERBOLA. 457 Let TR meet the curve, if possible. Draw FG, cutting TB at Q. Then OF' = OF; and QG=QF. §976 .-. F'G is II to TR. § 189 Therefore, F'G and TB cannot intersect. § 103 But if TB meets the curve, the point of contact must be at the intersection oi. F'G and TB. § 978 Therefore, TB does not meet the curve. q.e d 982. CoK. 1. The line FG is tangent to the auxiliary circle at Q. For FG is J- to OB. § 976 Therefore, Q lies on the auxiliary circle. § 977 Hence, FG touches the auxiliary circle at Q. § 253 983. CoE. 2. FQ is equal to the semi-conjugate axis h. For FQ" = 'OF^ - '0Q\ § 372 and 5^ = c^ - a". § 957 But 0F=c, and OQ = a. Therefore, FQ = h. 984. CoK. 3. If the tangent to the curve at A meets the asymptote OB at B, then AB — b. For A OAB = A0QF. § 142 Therefore, AB = FQ = h. 985. Dbf. a perpendicular to a tangent erected at the point of contact is called a normal. 986. Def. The terms subtangent and subnormal are used in the hyperbola in the same sense as in the ellipse. § 934 468 BOOK IX. CONIC SECTIONS. Pkoposition XXXIV. Theorem. 987. The section of a right circular cone made hy a plane that cuts both nappes of the cone is an hyperbola. Let a plane cut the lower nappe of the cone in the curve PAQ, and the upper nappe in the curve P'A'Q'. To prove that PAQ and JP'A'Q' are the two branches of an hyperbola. Proof. The plane passed through the axis of the cone per- pendicular to the secant plane cuts the surface of the cone in the elements BS, CS (prolonged through S), and the secant plane in the line -NN'- Describe the © 0, 0', tangent to B8, CS, NN'. Let the points of contact be D, H, F, and D', W, F', respectively. THE HYPERBOLA. 459 Turn BSC and the © and 0' about the axis of the cone. BS and CS will generate the surfaces of the two nappes of a right circular cone; and the © 0, 0' will generate spheres which touch the cone in the © DKH, lyK'W, and the secant plane in the points F, F'. Let P be any point on the curve. Draw PF and PF' ; and draw P*S,- which touches the ® DKH, D'K'IP, at the points E,K\ Now PF and PK are tangents to the sphere from the point P. Therefore, PF = PK. § 261 Also PF' = PK. § 261 Hence, PF' - PF = PK - PK Ax. 3 = KK, a constant quantity. § 716 Therefore, the curve is an hyperbola with the points F and F' for foci. § 945 Q. E.D. TABLE OF FORMULAS. PLANE FIGURES. h = lower base. b' = upper base. NOTATION. P = perimeter. h = altitude. R = radius of circle. D = diameter of circle. C = circumference of circle. r = apothem of regular polygon. a,b,c = sides of triangle. s = ^{a + b + c). p = perpendicular of triangle. m, n = segments of third side of triangle adjacent to sides b and a, respectively. S = area. tt = 3.1416. POBMUIjAS. ine Values. PAGE Right triangle, P = c X Tn; p^ = VI X n b'': a'^ = m : n b"^ : c^ : : m : c a^ + b^ = o'' . aP- = = c X n . . 160 . 160 . 161 . 161 . 162 460 TABLE OF FORMULAS. 461 Any triangle, a" = b'^ + o''±2c x m . 163, 164 Altitude of triangle on side a, 2 I A = - Vs (s - a) (s - i) (s - c) 180 Median of triangle on side a, m, = \ V2(J2 + c2)_a2 . . 180 Bisector of triangle on side a, t = b + c Radius of circnmscribed circle, R = ^bcs (s — a) abc 4 Vs (s — a) (s — b)(s — c) Circumference of circle, C = 2 wR Areas. Rectangle, Square, Parallelogram, Triangle, tt S=bxh . S=b^ . S=bx h S=ibxh . S=^s(s-a)(s- b) (s - c) abo S = 4iJ Equilateral triangle, '^ — T "^ Trapezoid, Regular polygon. Circle, a Sector, S=ih'(b + b') S=irXF . S=iBX C . (S = ^ -B X arc 181 181 222 222 187 187 188 189 208 208 208 190 223 224 224 224 462 TABLE OF FORMULAS. POLYHEDRONS, CYLINDERS, AND CONES. NOTATION. S = lateral area. V = volume. ^ = lateral edge ; element. i?" = altitude. P = perimeter of right section (Prisms and Cylinders). jj = perimeter of upper base. L = slant height. P = perimeter of lower base. B = lower base. c = circumference of upper base, b = upper base. C = circumference of lower base. T = total area, r = radius of upper base. M= area of mid-section. J? = radius of lower base. a, b, G = dimensions of paraUelopiped. FOEMULAS. Prisms and Parallelepipeds. PAOE S=EX P 293 V=B X R . ... 303, 305 V= a X b X c (Kectangular ParaUelopiped) . 302 Pyramids. S = iLX P (Regular Pyramid) . . .309 r=lBxH 316 Frustums of Pyramids.